The triumphs of general relativity

The ascent to greatness, however steep and dangerous, may entertain an active spirit with the consciousness and exercise of its own power: but the possession of a throne could never yet afford a lasting satisfaction to an ambitious mind. Edward Gibbon (1737-1794)
In our journey through general relativity's intellectual landscape, we have now finished a long, arduous, uphill trek and reached the mountain summit, on the top of which is engraved the Einstein field equation, which we write here in SI units as
(14.1) G = 8 π G c 4 T (14.1) G = 8 π G c 4 T {:(14.1)G=(8pi G)/(c^(4))T:}\begin{equation*} \boldsymbol{G}=\frac{8 \pi G}{c^{4}} \boldsymbol{T} \tag{14.1} \end{equation*}(14.1)G=8πGc4T
After all the hard effort expended in the steep climb of the previous chapters, discovering the constituents of the equation, their meaning and how they fit together, it's time to enjoy a well-earned breather to take in the view! Therefore, in this chapter we will review some of the triumphs of the theory, that is, areas where general relativity makes accurate predictions and provides convincing explanations of physical behaviour. All of the ideas outlined here will be discussed in far more detail throughout the remainder of the book.

14.1 Weak fields and the Newtonian limit

In our search for a relativistic theory of gravity, one thing we required in Chapter 6 was that the Einstein equation reproduces the results of Newton's theory in the non-relativistic limit. 1 1 ^(1){ }^{1}1 Our first task is to show that this is indeed the case. By 'non-relativistic' we usually mean the limit of velocities | u | | u | | vec(u)||\vec{u}||u| that are small compared to c c ccc, and so it makes sense to restore the factors of c c ccc in this section. More precisely, we find relativistic corrections are not required when | u | 2 / c 2 1 | u | 2 / c 2 1 | vec(u)|^(2)//c^(2)≪1|\vec{u}|^{2} / c^{2} \ll 1|u|2/c21 which means that γ = ( 1 | u | 2 / c 2 ) 1 2 1 γ = 1 | u | 2 / c 2 1 2 1 gamma=(1-|( vec(u))|^(2)//c^(2))^(-(1)/(2))~~1\gamma=\left(1-|\vec{u}|^{2} / c^{2}\right)^{-\frac{1}{2}} \approx 1γ=(1|u|2/c2)121. Since in a flat frame the relativistic velocity has components u μ = ( γ c , γ u i ) u μ = γ c , γ u i u^(mu)=(gamma c,gammau^(i))u^{\mu}=\left(\gamma c, \gamma u^{i}\right)uμ=(γc,γui), we also expect that u 0 u i u 0 u i u^(0)≫u^(i)u^{0} \gg u^{i}u0ui in this limit.
In a weak gravitational field, the acceleration due to gravity is not expected to lead to particles attaining a high velocity compared to c c ccc. Another way to say this using dimensional analysis is that small | u | | u | | vec(u)||\vec{u}||u| is equivalent to small G M / r G M / r GM//rG M / rGM/r, where G G GGG is the gravitational constant for a

14.1 Weak fields and the Newtonian limit
151 14.2 Gravitational waves 153 14.3 Stars, trajectories, and orbits or-
155 14.4 Cosmology 156 Chapter summary 156
1 1 ^(1){ }^{1}1 The consequences of taking the weakfield limit are discussed in more detail in Chapter 45.
2 2 ^(2){ }^{2}2 Factors of c c ccc can be restored here either using straightforward dimensional analysis, or by consulting the table in Chapter 0 .
3 3 ^(3){ }^{3}3 Because pressure is due to particles with kinetic energy 1 2 m v 2 1 2 m v 2 (1)/(2)mv^(2)\frac{1}{2} m v^{2}12mv2 bouncing off the container walls, the energy density ρ v 2 ρ v 2 ~~rhov^(2)\approx \rho v^{2}ρv2 and this, apart from a numerical factor, gives the pressure.
4 4 ^(4){ }^{4}4 This is the trace reversal of h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν (see the previous chapter), with the prop erty h ¯ = η μ ν h ¯ μ ν = h h ¯ = η μ ν h ¯ μ ν = h bar(h)=eta^(mu nu) bar(h)_(mu nu)=-h\bar{h}=\eta^{\mu \nu} \bar{h}_{\mu \nu}=-hh¯=ημνh¯μν=h.
source of mass M M MMM at distance r r rrr from a test particle. The non-relativistic limit, therefore, is the limit that ( G M / r c ) 2 1 ( G M / r c ) 2 1 (GM//rc)^(2)≪1(G M / r c)^{2} \ll 1(GM/rc)21, and we will use this condition to define what we mean by a weak-field limit.

Example 14.1

Consider replacing the factors of c c ccc in the components of the energy-momentum tensor for a perfect fluid. In a frame in which the elements of the fluid move with a velocity with components u μ u μ u^(mu)u^{\mu}uμ, we have 2 T μ ν = ( ρ + p / c 2 ) u μ u ν + p g μ ν 2 T μ ν = ρ + p / c 2 u μ u ν + p g μ ν ^(2)T^(mu nu)=(rho+p//c^(2))u^(mu)u^(nu)+pg^(mu nu){ }^{2} T^{\mu \nu}=\left(\rho+p / c^{2}\right) u^{\mu} u^{\nu}+p g^{\mu \nu}2Tμν=(ρ+p/c2)uμuν+pgμν. Since 3 3 ^(3){ }^{3}3 the pressure p ρ v 2 p ρ v 2 p~~rhov^(2)p \approx \rho v^{2}pρv2, where ρ ρ rho\rhoρ is the mass density, then ρ p / c 2 ρ p / c 2 rho≫p//c^(2)\rho \gg p / c^{2}ρp/c2 in the weak-field limit, and so we have T μ ν ρ u μ u ν T μ ν ρ u μ u ν T^(mu nu)~~rhou^(mu)u^(nu)T^{\mu \nu} \approx \rho u^{\mu} u^{\nu}Tμνρuμuν and, since u 0 u i u 0 u i u^(0)≫u^(i)u^{0} \gg u^{i}u0ui, we also have T 00 T i i T 00 T i i T^(00)≫T^(ii)T^{00} \gg T^{i i}T00Tii.
In the weak-field limit, the gravitational metric field can be modelled as making a small perturbation to the Minkowski metric for flat space. We then have
(14.2) g μ ν = η μ ν + h μ ν (14.2) g μ ν = η μ ν + h μ ν {:(14.2)g_(mu nu)=eta_(mu nu)+h_(mu nu):}\begin{equation*} g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu} \tag{14.2} \end{equation*}(14.2)gμν=ημν+hμν
with | h μ ν | 1 h μ ν 1 |h_(mu nu)|≪1\left|h_{\mu \nu}\right| \ll 1|hμν|1. Note that, as we shall see later, h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν here are not actually the components of a tensor, but we raise and lower all components with η μ ν η μ ν eta_(mu nu)\eta_{\mu \nu}ημν. Equation 14.2 can be inserted into the Einstein equation to deduce an equation for h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν in terms of the energy-momentum tensor T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν. It turns out that the equation we need comes out most cleanly in terms of h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν, an object related 4 4 ^(4){ }^{4}4 to h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν defined by
(14.3) h ¯ μ ν = h μ ν 1 2 η μ ν h (14.3) h ¯ μ ν = h μ ν 1 2 η μ ν h {:(14.3) bar(h)_(mu nu)=h_(mu nu)-(1)/(2)eta_(mu nu)h:}\begin{equation*} \bar{h}_{\mu \nu}=h_{\mu \nu}-\frac{1}{2} \eta_{\mu \nu} h \tag{14.3} \end{equation*}(14.3)h¯μν=hμν12ημνh
and h h hhh is the trace h = η μ ν h μ ν h = η μ ν h μ ν h=eta^(mu nu)h_(mu nu)h=\eta^{\mu \nu} h_{\mu \nu}h=ημνhμν. We then find that
(14.4) 2 h ¯ μ ν = 16 π G c 4 T μ ν (14.4) 2 h ¯ μ ν = 16 π G c 4 T μ ν {:(14.4)del^(2) bar(h)_(mu nu)=-(16 pi G)/(c^(4))T_(mu nu):}\begin{equation*} \partial^{2} \bar{h}_{\mu \nu}=-\frac{16 \pi G}{c^{4}} T_{\mu \nu} \tag{14.4} \end{equation*}(14.4)2h¯μν=16πGc4Tμν

Example 14.2

Let's evaluate eqn 14.4 in the weak-field limit for T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν which means that, setting c = 1 , 5 c = 1 , 5 c=1,^(5)c=1,{ }^{5}c=1,5 the only appreciable term is T 00 T 00 ρ T 00 T 00 ρ T_(00)~~T^(00)~~rhoT_{00} \approx T^{00} \approx \rhoT00T00ρ. We deduce that 2 h ¯ 00 = 2 h ¯ 00 = -del^(2) bar(h)_(00)=-\partial^{2} \bar{h}_{00}=2h¯00= 16 π G T 00 = 16 π G ρ 16 π G T 00 = 16 π G ρ 16 pi GT_(00)=16 pi G rho16 \pi G T_{00}=16 \pi G \rho16πGT00=16πGρ, but we note that 2 = 2 / t 2 + 2 2 = 2 / t 2 + 2 del^(2)=-del^(2)//delt^(2)+grad^(2)\partial^{2}=-\partial^{2} / \partial t^{2}+\nabla^{2}2=2/t2+2 and so the time dependence can be neglected (because all speeds are much less than c c ccc ). Hence, we have
(14.6) 2 h ¯ 00 = 16 π G ρ (14.6) 2 h ¯ 00 = 16 π G ρ {:(14.6)grad^(2) bar(h)_(00)=-16 pi G rho:}\begin{equation*} \nabla^{2} \bar{h}_{00}=-16 \pi G \rho \tag{14.6} \end{equation*}(14.6)2h¯00=16πGρ
with the other components of h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν obeying a Laplace equation (i.e. with no source), so will be zero (assuming that these terms should fall off to zero at infinity anyway), Comparing with the Newtonian result 2 Φ = 4 π G ρ 2 Φ = 4 π G ρ grad^(2)Phi=4pi G rho\nabla^{2} \Phi=4 \pi G \rho2Φ=4πGρ we deduce that h ¯ 00 = 4 Φ h ¯ 00 = 4 Φ bar(h)_(00)=-4Phi\bar{h}_{00}=-4 \Phih¯00=4Φ with all other components of h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν equal to zero. Hence, the trace of h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν is h ¯ = 4 Φ h ¯ = 4 Φ bar(h)=4Phi\bar{h}=4 \Phih¯=4Φ and we can therefore deduce the perturbation h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν to the metric using
(14.7) h μ ν = h ¯ μ ν 1 2 η μ ν h ¯ (14.7) h μ ν = h ¯ μ ν 1 2 η μ ν h ¯ {:(14.7)h_(mu nu)= bar(h)_(mu nu)-(1)/(2)eta_(mu nu) bar(h):}\begin{equation*} h_{\mu \nu}=\bar{h}_{\mu \nu}-\frac{1}{2} \eta_{\mu \nu} \bar{h} \tag{14.7} \end{equation*}(14.7)hμν=h¯μν12ημνh¯
yielding
h μ ν = ( 4 Φ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) 1 2 ( 4 Φ ) ( 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ) (14.8) = ( 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ ) . h μ ν = 4 Φ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 ( 4 Φ ) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 (14.8) = 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ . {:[h_(mu nu)=([-4Phi,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0])-(1)/(2)(4Phi)([-1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1])],[(14.8)=([-2Phi,0,0,0],[0,-2Phi,0,0],[0,0,-2Phi,0],[0,0,0,-2Phi]).]:}\begin{align*} h_{\mu \nu} & =\left(\begin{array}{cccc} -4 \Phi & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)-\frac{1}{2}(4 \Phi)\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \\ & =\left(\begin{array}{cccc} -2 \Phi & 0 & 0 & 0 \\ 0 & -2 \Phi & 0 & 0 \\ 0 & 0 & -2 \Phi & 0 \\ 0 & 0 & 0 & -2 \Phi \end{array}\right) . \tag{14.8} \end{align*}hμν=(4Φ000000000000000)12(4Φ)(1000010000100001)(14.8)=(2Φ00002Φ00002Φ00002Φ).
The full metric (Minkowski plus the small perturbation) then becomes
g μ ν = η μ ν + h μ ν = ( 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ) + ( 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ ) ( 14.9 ) g μ ν = η μ ν + h μ ν = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 + 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ 0 0 0 0 2 Φ ( 14.9 ) g_(mu nu)=eta_(mu nu)+h_(mu nu)=([-1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1])+([-2Phi,0,0,0],[0,-2Phi,0,0],[0,0,-2Phi,0],[0,0,0,-2Phi])*(14.9)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}=\left(\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)+\left(\begin{array}{cccc} -2 \Phi & 0 & 0 & 0 \\ 0 & -2 \Phi & 0 & 0 \\ 0 & 0 & -2 \Phi & 0 \\ 0 & 0 & 0 & -2 \Phi \end{array}\right) \cdot(14.9)gμν=ημν+hμν=(1000010000100001)+(2Φ00002Φ00002Φ00002Φ)(14.9)
Writing the metric in terms of the line element gives
(14.10) d s 2 = g μ ν d x μ d x ν = ( 1 + 2 Φ ) d t 2 + ( 1 2 Φ ) ( d x 2 + d y 2 + d z 2 ) (14.10) d s 2 = g μ ν d x μ d x ν = ( 1 + 2 Φ ) d t 2 + ( 1 2 Φ ) d x 2 + d y 2 + d z 2 {:(14.10)ds^(2)=g_(mu nu)dx^(mu)dx^(nu)=-(1+2Phi)dt^(2)+(1-2Phi)(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} \mathrm{d} s^{2}=g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}=-(1+2 \Phi) \mathrm{d} t^{2}+(1-2 \Phi)\left(\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2}\right) \tag{14.10} \end{equation*}(14.10)ds2=gμνdxμdxν=(1+2Φ)dt2+(12Φ)(dx2+dy2+dz2)
finally proving the result given in eqn 5.22 .

14.2 Gravitational waves

The metric field, just like the electromagnetic field, can host waves in space and time. These gravitational waves cause periodic distortions in the metric field that can tell masses how to move. Their effects are, however, very small, and escaped detection until 2015 when a burst of waves lasting 150 ms were detected by the LIGO gravitational wave detectors. 6 6 ^(6){ }^{6}6 This burst was deduced to have resulted from the collision of two black holes: an event with such an immense output of power that the gravitational waves could finally pass the threshold for detection.

Example 14.3

We saw in Chapter 11 that the effect of the tidal forces caused by a massive planet is to distort a spherical distribution of mass into an ellipsoid of the same volume. The distribution is stretched in the direction of the planet and contracted in the other two orthogonal directions. The reason behind this conservation of volume can actually be traced back to the Ricci tensor, whose components are R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν. The Einstein equation given in terms of these tensor components reads
(14.11) R μ ν ( x ) 1 2 g μ ν R ( x ) = 8 π G c 4 T μ ν ( x ) (14.11) R μ ν ( x ) 1 2 g μ ν R ( x ) = 8 π G c 4 T μ ν ( x ) {:(14.11)R_(mu nu)(x)-(1)/(2)g_(mu nu)R(x)=(8pi G)/(c^(4))T_(mu nu)(x):}\begin{equation*} R_{\mu \nu}(x)-\frac{1}{2} g_{\mu \nu} R(x)=\frac{8 \pi G}{c^{4}} T_{\mu \nu}(x) \tag{14.11} \end{equation*}(14.11)Rμν(x)12gμνR(x)=8πGc4Tμν(x)
This is a local equation (hence, the dependence on the position in spacetime x x xxx ), and so, if T ( x ) T ( x ) T(x)\boldsymbol{T}(x)T(x) vanishes at a point, then so does the right-hand side of the equation. This will be the case in vacuum, where there are no sources of energy-momentum. This, in turn, implies that R μ ν ( x ) R μ ν ( x ) R_(mu nu)(x)R_{\mu \nu}(x)Rμν(x) vanishes in the vacuum, and provides an explanation for the volume-preserving property of the tidal forces from the planet: in order for R μ ν ( x ) R μ ν ( x ) R_(mu nu)(x)R_{\mu \nu}(x)Rμν(x) to vanish, 7 7 ^(7){ }^{7}7 the tidal forces must have positive parts (that cause contraction of the ellipse) exactly cancelling the negative parts (that stretch out the ellipse in the orthogonal direction). 8 8 ^(8){ }^{8}8
Since, in free space, gravitational waves are also governed by R μ ν ( x ) = 0 R μ ν ( x ) = 0 R_(mu nu)(x)=0R_{\mu \nu}(x)=0Rμν(x)=0, there is a similar balance between stretching (negative curvature) and contraction (positive curvature) owing to the existence of the wave. In this case, there is a stretching effect and a compressive effect in one orthogonal direction (not two directions, as we had for the planet example above). To achieve the cancellation, these must be equal and opposite.
From the last example, we can draw a picture of some typical gravitational waves. These are shown in terms of their effect on a circle of
\curvearrowright Chapter 45 contains a more detailed derivation of the equations in Section 14.1, and further discussion of the weakfield limit.
6 6 ^(6){ }^{6}6 LIGO stands for Laser Interferometer Gravitational-Wave Observatory.
7 7 ^(7){ }^{7}7 Recall that R μ ν ( x ) = R α μ α ν ( x ) R μ ν ( x ) = R α μ α ν ( x ) R_(mu nu)(x)=R^(alpha)_(mu alpha nu)(x)R_{\mu \nu}(x)=R^{\alpha}{ }_{\mu \alpha \nu}(x)Rμν(x)=Rαμαν(x), so can be thought of as a sort of average of the Riemann curvature at a point, retaining the parts of the curvature that cause volumes to shrink.
8 8 ^(8){ }^{8}8 This makes sense if we refer back to Chapter 11, where we saw that the compressive forces were half as large as the stretching ones, but since the compression occurs in two directions, we achieve the cancellation we have claimed. The properties of the Ricci tensor are discussed in more detail in Chapter 35.
Fig. 14.1 (a) A circle of test masses in flat space. (b) The effect of a gravitational wave on the circle of test masse at two different times, corresponding to half a period of the gravitational wave. (c) The same as (b) but with a different polarization.
9 9 ^(9){ }^{9}9 The polarization of an electromagnetic wave can be linear or circular, or a superposition expressed in a linear or circular basis.
10 10 ^(10){ }^{10}10 See Chapter 42.
11 11 ^(11){ }^{11}11 For example, the fact that h μ ν h μ ν h_(mu nu)h_{\mu \nu}hμν is symmetrical immediately reduces the potential number of independent components from sixteen to ten.
masses [Fig. 14.1(a)] in Fig. 14.1(b) and (c) for two different polarizations. In (b), for example, we see the effect at some point in spacetime of expansion and contraction of the circle into an ellipse. Half a wavelength further along the direction of propagation of the wave, the stretching and compressive forces have swapped over, causing the circle of masses to be distorted in the opposite sense.
In order to understand where the waves come from, we must look at the behaviour of the metric field, and in particular we can look at the perturbation to the Minkowski metric h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν introduced earlier in the chapter. In the absence of any matter, the components of the energymomentum tensor vanish, and eqn 14.4 reduces to
(14.12) 2 h ¯ μ ν = 0 (14.12) 2 h ¯ μ ν = 0 {:(14.12)-del^(2) bar(h)_(mu nu)=0:}\begin{equation*} -\partial^{2} \bar{h}_{\mu \nu}=0 \tag{14.12} \end{equation*}(14.12)2h¯μν=0
This is a wave equation for the components h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν.

Example 14.4

A wave equation for a scalar field ϕ ( t , x ) ϕ ( t , x ) phi(t, vec(x))\phi(t, \vec{x})ϕ(t,x) looks like 2 ϕ = 0 2 ϕ = 0 del^(2)phi=0\partial^{2} \phi=02ϕ=0, or, in component form
(14.13) 1 c 2 2 ϕ t 2 + 2 ϕ x 2 + 2 ϕ y 2 + 2 ϕ z 2 = 0 . (14.13) 1 c 2 2 ϕ t 2 + 2 ϕ x 2 + 2 ϕ y 2 + 2 ϕ z 2 = 0 . {:(14.13)-(1)/(c^(2))(del^(2)phi)/(delt^(2))+(del^(2)phi)/(delx^(2))+(del^(2)phi)/(dely^(2))+(del^(2)phi)/(delz^(2))=0.:}\begin{equation*} -\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}+\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0 . \tag{14.13} \end{equation*}(14.13)1c22ϕt2+2ϕx2+2ϕy2+2ϕz2=0.
The solutions to this equation of motion are plane waves such as ϕ ( t , x ) = C e i k x = ϕ ( t , x ) = C e i k x = phi(t, vec(x))=Ce^(ik*x)=\phi(t, \vec{x})=C \mathrm{e}^{\mathrm{i} k \cdot \boldsymbol{x}}=ϕ(t,x)=Ceikx= C e i ( ω t k x ) C e i ( ω t k x ) Ce^(-i(omega t- vec(k)* vec(x)))C \mathrm{e}^{-\mathrm{i}(\omega t-\vec{k} \cdot \vec{x})}Cei(ωtkx), with ω = c | k | ω = c | k | omega=c| vec(k)|\omega=c|\vec{k}|ω=c|k| and C C CCC a constant. The waves are excitations of the scalar field, requiring a non-zero, but finite energy to come into existence. Similarly, a wave equation for a 1 -form field with components A μ A μ A_(mu)A_{\mu}Aμ can be written as 2 A μ = 0 2 A μ = 0 del^(2)A_(mu)=0\partial^{2} A_{\mu}=02Aμ=0 with solutions
(14.14) A μ = Re [ C μ e i ( ω t k x ) ] (14.14) A μ = Re C μ e i ( ω t k x ) {:(14.14)A_(mu)=Re[C_(mu)e^(-i(omega t- vec(k)* vec(x)))]:}\begin{equation*} A_{\mu}=\operatorname{Re}\left[C_{\mu} \mathrm{e}^{-\mathrm{i}(\omega t-\vec{k} \cdot \vec{x})}\right] \tag{14.14} \end{equation*}(14.14)Aμ=Re[Cμei(ωtkx)]
The relationship between the different components give us the idea of the polarization of the wave. 9 9 ^(9){ }^{9}9 In the absence of sources, we can write such a wave equation for the electromagnetic field A ( x ) A ¯ ( x ) bar(A)(x)\overline{\boldsymbol{A}}(x)A(x), from which the more familiar electric and magnetic fields can be extracted using 10 10 ^(10){ }^{10}10
(14.15) E = A 0 A t , B = × A (14.15) E = A 0 A t , B = × A {:(14.15) vec(E)=- vec(grad)A^(0)-(del( vec(A)))/(del t)","quad vec(B)= vec(grad)xx vec(A):}\begin{equation*} \vec{E}=-\vec{\nabla} A^{0}-\frac{\partial \vec{A}}{\partial t}, \quad \vec{B}=\vec{\nabla} \times \vec{A} \tag{14.15} \end{equation*}(14.15)E=A0At,B=×A
A wave equation for the components h ¯ μ ν h ¯ μ ν bar(h)_(mu nu)\bar{h}_{\mu \nu}h¯μν, describing the weak gravitational field, means that this field can also support wave-like excitations, just like those in the last example. These gravitational waves have the potential to move masses in periodic motions. The solutions to the wave equation look like
(14.16) h ¯ μ ν = Re [ A μ ν e i k x ] (14.16) h ¯ μ ν = Re A μ ν e i k x {:(14.16) bar(h)_(mu nu)=Re[A_(mu nu)e^(ik*x)]:}\begin{equation*} \bar{h}_{\mu \nu}=\operatorname{Re}\left[A_{\mu \nu} \mathrm{e}^{\mathrm{i} \boldsymbol{k} \cdot \boldsymbol{x}}\right] \tag{14.16} \end{equation*}(14.16)h¯μν=Re[Aμνeikx]
A cause of complication here is that the field has two indices, implying sixteen components in (3+1)-dimensional spacetime. Although these components are not all independent (so several are identical), 11 11 ^(11){ }^{11}11 this feature leads to the orthogonal stretching and compressive forces along with the polarizations shown in Fig. 14.1.

14.3 Stars, trajectories, and orbits

Not too many solutions of the Einstein equation are known, but one of the most useful exact solutions is the Schwarzschild metric, which describes the geometry of spacetime near a static, spherically symmetric distribution of mass (with total mass M M MMM ). The line element of his metric is written in spherical coordinates as
d s 2 = ( 1 2 G M r c 2 ) c 2 d t 2 + ( 1 2 G M r c 2 ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 G M r c 2 c 2 d t 2 + 1 2 G M r c 2 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-(1-(2GM)/(rc^(2)))c^(2)dt^(2)+(1-(2GM)/(rc^(2)))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))\mathrm{d} s^{2}=-\left(1-\frac{2 G M}{r c^{2}}\right) c^{2} \mathrm{~d} t^{2}+\left(1-\frac{2 G M}{r c^{2}}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)ds2=(12GMrc2)c2 dt2+(12GMrc2)1 dr2+r2( dθ2+sin2θ dϕ2).
(14.17)
In the free space outside of the mass distribution, we have no local sources of energy momentum, and so T = 0 T = 0 T=0\boldsymbol{T}=0T=0. The line element then solves the Einstein equation G = 0 G = 0 G=0\boldsymbol{G}=0G=0. The motion of masses in this geometry allows us to see the affects of general relativity on the planets in our solar system. There are several famous solar-system-based tests of general relativity, the three known as the 'classical' ones are listed in the margin. Here's we'll concentrate on the first of these.
One triumph of Newton's theory was the prediction of elliptical orbits of planets around the Sun. General relativity provides a richer array of orbits and motions of massive particles in a gravitational field. We can describe the motion as that of a particle in an effective potential V eff ( r ) V eff  ( r ) V_("eff ")(r)V_{\text {eff }}(r)Veff (r). We write
(14.18) E = 1 2 ( d r d τ ) 2 + V eff ( r ) (14.18) E = 1 2 d r d τ 2 + V eff ( r ) {:(14.18)E=(1)/(2)(((d)r)/((d)tau))^(2)+V_(eff)(r):}\begin{equation*} \mathcal{E}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\mathrm{eff}}(r) \tag{14.18} \end{equation*}(14.18)E=12( dr dτ)2+Veff(r)
where the effective potential for a particle is given by
(14.19) V eff ( r ) = G M r c 2 + L ~ 2 2 r 2 c 2 G M L ~ 2 r 3 c 4 (14.19) V eff ( r ) = G M r c 2 + L ~ 2 2 r 2 c 2 G M L ~ 2 r 3 c 4 {:(14.19)V_(eff)(r)=-(GM)/(rc^(2))+( tilde(L)^(2))/(2r^(2)c^(2))-(GM tilde(L)^(2))/(r^(3)c^(4)):}\begin{equation*} V_{\mathrm{eff}}(r)=-\frac{G M}{r c^{2}}+\frac{\tilde{L}^{2}}{2 r^{2} c^{2}}-\frac{G M \tilde{L}^{2}}{r^{3} c^{4}} \tag{14.19} \end{equation*}(14.19)Veff(r)=GMrc2+L~22r2c2GML~2r3c4
where L ~ L ~ tilde(L)\tilde{L}L~ refers to the angular momentum per unit mass of the orbiting particle. The first two terms are also found in Newtonian gravitation; the final term is the result of general relativity, providing a correction to the motion.
Some quite general allowed trajectories in this potential are shown in Fig. 14.2. They depend on the value of the effective potential, which is dependent on the (conserved) value of angular momentum of the particle. The most striking of these is the precession of the near-elliptical orbits shown in Fig. 14.3, by which we mean the apparent rotation of the elliptical orbit of a planet about the sun. The accurate prediction of the perihelion shift (i.e. the rate of this rotation) of the planet mercury due to this precession effect was historically one of the most significant of the early triumphs of Einstein's theory. The calculated value of precession is approximately 43.0 arc seconds per century, a small but observable quantity, which agrees very well with experiment.
In contrast to the Newtonian theory, light is also predicted to be bent by gravitational fields in general relativity. A similar approach to that outlined above for massive particles, yields an effective potential for light rays. This leads to the prediction of gravitational-lensing effects.
The three classical solar system tests of relativity are:
  • Precession of the perihelion of mercury;
  • Bending of light by the sun;
  • Gravitational redshift.

    (c) V eff ( r ) V eff  ( r ) V_("eff ")( vec(r))V_{\text {eff }}(\vec{r})Veff (r)

Fig. 14.2 Some allowed trajectories in the relativistic potential. (a) Circular; (b) an encounter with the star leading to scattering; (c) a spiral into the star.
Fig. 14.3 The precessing orbit in the relativistic potential. The motion is bounded at the points shown on the left, represented by the dotted lines on the right.
\curvearrowright Part IV of the book discusses orbits, stars and blackholes, further developing the ideas outlined in Section 14.3.
\curvearrowright Part III of the book, coming up next, is concerned with cosmology, giving much more detail on the subject outlined in Section 14.4.
The most extreme gravitational effects occur in this geometry close to a source of the gravitational field which is very dense. For particularly dense mass distributions, we obtain the prediction of a black hole, which is a region of spacetime where the gravitational field is so strong that not even light can escape its clutches.

14.4 Cosmology

In addition to treating the physics of the trajectories of individual bodies, general relativity can describe the spacetime structure of the entire Universe. This is the field of cosmology, which gives us an insight into the origin and fate of our Universe.
Many model universes have a geometry described by the RobertsonWalker metric, with line element
(14.20) d s 2 = c 2 d t 2 + a ( t ) 2 [ d r 2 1 k r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] , (14.20) d s 2 = c 2 d t 2 + a ( t ) 2 d r 2 1 k r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 , {:(14.20)ds^(2)=-c^(2)dt^(2)+a(t)^(2)[((d)r^(2))/(1-kr^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]",":}\begin{equation*} \mathrm{d} s^{2}=-c^{2} \mathrm{~d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-k r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right], \tag{14.20} \end{equation*}(14.20)ds2=c2 dt2+a(t)2[ dr21kr2+r2( dθ2+sin2θ dϕ2)],
where the parameter k = 1 , 0 k = 1 , 0 k=1,0k=1,0k=1,0 or -1 , depending on the curvature of spacetime. This line element solves the Einstein equation and describes universes with geometries that, when filled with different sorts of matter, have distinct origins and fates. Perhaps the most significant feature of these model universes is that the spatial parts of the metric vary with the time coordinate t t ttt, giving us a notion of cosmic evolution. In fact, the mechanics of spacetime can be determined via equations of motion for a ( t ) a ( t ) a(t)a(t)a(t), which is a quantity that functions as a sort of effective radius of the Universe. Notably, many of these model universes start from a Big Bang: an initial state where the whole universe is shrunk to a point. (The universe then undergoes a period of expansion.) The Big Bang represents an example of a singularity, which is a point in spacetime where our classical field theories, such as general relativity, break down.
Clearly, there is a lot to unpack here and it is to cosmology that we turn in the next part of the book.

Chapter summary

  • The Einstein equation matches the Newtonian theory in the weakfield limit.
  • Gravitational waves are excitations of the gravitational field.
  • Newtonian theory predicted the existence of elliptical orbits in central potentials, but general relativity leads to precession of elliptical-like orbits.
  • Cosmology, the application of the Einstein equation to the entire Universe, results in descriptions of the origin and fate of the Universe.

Part III

Cosmology

Having formulated a theory of gravity, we are now going to use it on the largest conceivable problem: the origin and fate of our Universe.
  • In Chapter 15, we introduce the ideas behind the field of cosmology and investigate two model universes.
  • In Chapter 16, we investigate the possible geometries describing homogeneous, isotopic spacetimes with constant curvature and in Chapter 17 we fill these with matter.
  • We apply all of these ideas in Chapter 18 to a number of different model universes.
  • In the final chapter of Part III (Chapter 19), we study some very general ideas that allow us to think about infinity in the context of the geometry of spacetime.

15

15.1 The cosmological principle

15.2 The Hubble flow 160

15.3 Cosmic time 162
15.4 Universe 0: an empty universe 163
15.5 Universe 1: flat and expanding
1 1 ^(1){ }^{1}1 In addition to gravity, these are electromagnetism, the weak nuclear force and the strong nuclear force

An introduction to cosmology

To see a World in a Grain of SandAnd a Heaven in a Wild FlowerHold Infinity in the palm of your handAnd Eternity in an hourWilliam Blake (1757-1827)I don't pretend to understand the Universe - it's a great deal bigger than I am ... People ought to be modester. Thomas Carlyle (1795-1881)

After several chapters spent building a field theory of gravitation, we are now ready to use it. Our first task will be the, admittedly ambitious, application of Einstein's equation to the entire Universe! Gravitation is the weakest of the four so-called fundamental interactions. 1 1 ^(1){ }^{1}1 However, unlike the strong and weak nuclear interactions, gravity is a long-range force. Moreover, unlike the (similarly long-range) electromagnetic interaction, gravitation is always attractive and so cannot cancel out in the same way that, owing to a balance in the number of positive and negative electric charges, electromagnetism frequently does. Gravitation is therefore always attractive, additive, and acts over large distances, causing it to be the dominant interaction in determining the large-scale behaviour of the Universe.
In this chapter, we begin an investigation of cosmology by applying Einstein's equation to describe models of the Universe in terms of the large-scale structure of spacetime. First, some good news: there are several solvable models of the Universe. In general, these have the property that the spatial parts of the metric field depend on the time coordinate, leading to the idea of cosmic evolution, a universe changing with time. Although this disturbed early relativists who expected a steady-state universe, we are now more comfortable with this idea. Second, the bad news: many of the universes appear to start, and some end, in a singularity. This is a point in spacetime that cannot be dealt with using classical field theory, which is designed to deal with smoothly evolving spacetime. A singularity is a point in spacetime where the known laws of physics break down, and that doesn't sound like a good thing. However, in later chapters we shall learn how to accommodate them. In any case, let's begin by setting out the nature of the problem and some first attempts at solutions.

15.1 The cosmological principle

The central idea of cosmology is to solve the Einstein field equation 2 2 ^(2){ }^{2}2
(15.1) R μ ν 1 2 g μ ν R = 8 π T μ ν Λ g μ ν , (15.1) R μ ν 1 2 g μ ν R = 8 π T μ ν Λ g μ ν , {:(15.1)R_(mu nu)-(1)/(2)g_(mu nu)R=8piT_(mu nu)-Lambdag_(mu nu)",":}\begin{equation*} R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R=8 \pi T_{\mu \nu}-\Lambda g_{\mu \nu}, \tag{15.1} \end{equation*}(15.1)Rμν12gμνR=8πTμνΛgμν,
for all of the matter in the Universe. 3 3 ^(3){ }^{3}3 A solution involves a complete description of spacetime, achieved by specifying the metric field at each point in space and time. 4 4 ^(4){ }^{4}4 Cosmology therefore promises to provide us with a complete picture of all space in the past, present, and future.
To make progress in cosmology, we need a simplifying idea to get us started. We start with the Copernican principle 5 5 ^(5){ }^{5}5 which states that the view of the Universe that we have from our vantage point on Earth is not special; the Universe would look the same from any other viewpoint, at this epoch in cosmological history. This idea is closely related to the cosmological principle, which is stated as follows:
The cosmological principle says that the Universe is spatially homogeneous and isotropic.
  • The property of spatial homogeneity means that the properties of the Universe are not a function of position in the Universe; wherever you are in the Universe, everything looks the same.
  • The property of spatial isotropy means that there are no special directions; in whichever direction you look from your vantage point in the Universe, everything looks the same.
Although it is a great simplifying assumption, the cosmological principle is clearly wrong. If you look up, you see the sky; if you look down you see the ground; the Universe doesn't look isotropic. What's more, it doesn't seem homogeneous, as we can change our environment from mountains to valleys simply by changing our spatial position, and they are clearly different. Even looking at larger length scales doesn't seem to help as, for example, the Solar System (which is filled with planets, asteroids, comets and with a large star in the centre) seems very different from the intergalactic medium (which is largely empty). The problem here is that we are still looking at too small a length scale and we need to broaden our horizons.

Example 15.1

In Example 0.3, we introduced the Astronomical Unit, the distance from the Earth to the Sun, which is 1.496 × 10 11 m 1.496 × 10 11 m 1.496 xx10^(11)m1.496 \times 10^{11} \mathrm{~m}1.496×1011 m. However, this is a relatively small distance compared to the light-year which is the distance light travels in a year, roughly 9.46 × 10 15 m 9.46 × 10 15 m 9.46 xx10^(15)m9.46 \times 10^{15} \mathrm{~m}9.46×1015 m or more than 60,000 Astronomical Units. Astrophysicists tend to use, more often, the parsec (abbreviated to pc), which is 3.26 light-years: 6 6 ^(6){ }^{6}6
(15.2) 1 pc = 3.086 × 10 16 m (15.2) 1 pc = 3.086 × 10 16 m {:(15.2)1pc=3.086 xx10^(16)m:}\begin{equation*} 1 \mathrm{pc}=3.086 \times 10^{16} \mathrm{~m} \tag{15.2} \end{equation*}(15.2)1pc=3.086×1016 m
2 2 ^(2){ }^{2}2 Reminder: In this part of the book, we will take G = c = 1 G = c = 1 G=c=1G=c=1G=c=1, unless otherwise indicated.
3 3 ^(3){ }^{3}3 It might be worth keeping in mind that our basic strategy in this part of the book will be to solve the Einstein equation G = κ T Λ g G = κ T Λ g G=kappa T-Lambda g\boldsymbol{G}=\kappa \boldsymbol{T}-\Lambda \boldsymbol{g}G=κTΛg, by matching the components of G G G\boldsymbol{G}G with those of the energy-momentum contributions from T T TTT and Λ g Λ g Lambda g\Lambda \boldsymbol{g}Λg.
4 4 ^(4){ }^{4}4 This is a feature of the spacetime view of cosmology: the behaviour of spacetime is completely predetermined, and is therefore, in a sense, unchanging. This statement about spacetime as a whole does not contradict our assertion above that the spatial parts of the metric depend on the time coordinate.
5 5 ^(5){ }^{5}5 Nicolaus Copernicus' (1473-1543) book De Revolutionibus described a heliocentric cosmology, removing the Earth from the centre of the Universe. Arthur Koestler (1905-1983) described the book as an 'all-time worst-seller', claiming it had not been widely read after its publication. The claim was disproved by Owen Gingerich (1930- ), as described in his The Book Nobody Read, after examination of almost every surviving copy of Copernicus book. The Copernican principle, in its modern cosmological form, was named by Hermann Bondi (1919-2005).
6 6 ^(6){ }^{6}6 A parsec is defined as the distance at which an astronomical unit subtends an angle of one arc second. Its name comes from parallax, the technique by which distances of nearby stars can be measured, and is essentially an abbreviation of 'parallax of one second'.
7 7 ^(7){ }^{7}7 This is like looking at a uniform grey rectangle on a computer screen; only by close examination with a magnifying glass will you notice that it is made up of tiny dots. Similarly, only by close examination of the Universe (on scales much less than 100 Mpc ) do you notice that it is actually made up of discrete galaxies, themselves composed of individual stars.
8 8 ^(8){ }^{8}8 You can think of this as the distance to a galaxy from the Earth, where the latter sits at the origin of the coordinate system.
9 9 ^(9){ }^{9}9 Edwin Hubble (1889-1953). Hubble's original measurement was nearly an or der of magnitude larger than this, but the value has been refined over the intervening period and we now believe we have it to within about 10 % 10 % 10%10 \%10% or so. Is it really a constant, taking the same value for all time? As we will see later, various cosmological models will show that it isn't, and so we will later refer to it as the Hubble parameter H ( t ) H ( t ) H(t)H(t)H(t).
The nearest star, Proxima Centauri, is 1.3 parsecs from us, and the brightest star in the night sky, Sirius, is 2.7 parsecs away. Thus, a parsec gives a rough estimate for the distance between stars in our Galaxy, the Milky Way. There are about 10 11 10 11 10^(11)10^{11}1011 stars in the Milky Way, shaped like a disc with a radius of something like 25 kpc which is around 0.3 kpc thick ( kpc means kiloparsec). The Andromeda Galaxy, the nearest large galaxy to the Milky Way, is about 770 kpc away from us. Thus, even on the scale of a megaparsec, the Universe looks quite lumpy and not at all homogeneous and isotropic. However, if you zoom out further, say to a scale of 100 Mpc (which is 3 × 10 24 m 3 × 10 24 m 3xx10^(24)m3 \times 10^{24} \mathrm{~m}3×1024 m, or more than thirteen orders of magnitude above the astronomical unit), the graininess and lumpiness of the Universe starts to disappear and the cosmological principle starts to hold.
Our theories of cosmology will therefore be constructed to understand the Universe on length scales longer than 100 Mpc 100 Mpc ~~100Mpc\approx 100 \mathrm{Mpc}100Mpc, and they will clearly not work on length scales shorter than this. Therefore, when we do cosmology we will be applying the Einstein field equations to a model Universe in which all the matter has been smeared out (the technical term is coarse-grained) so that it becomes a fluid of uniform density. 7 7 ^(7){ }^{7}7 The averaged-out mass content of the Universe is, in the absence of internal interactions, equivalent to the dust we discussed in Chapter 12. However, we shall often allow internal interactions too. In either case, the matter constitutes a perfect cosmological fluid (Chapter 12) that fills the Universe. We can allow the cosmic fluid to move, expand, and contract.

15.2 The Hubble flow

An important observational fact about the Universe was discovered by Edwin Hubble. From measurements of redshifts of emission lines in astrophysical objects we now know that there is a relationship between the velocity v v vec(v)\vec{v}v and position 8 r 8 r ^(8) vec(r){ }^{8} \vec{r}8r of an object in the Universe which is given by Hubble's law, which can be written as
(15.3) v = H 0 r (15.3) v = H 0 r {:(15.3) vec(v)=H_(0) vec(r):}\begin{equation*} \vec{v}=H_{0} \vec{r} \tag{15.3} \end{equation*}(15.3)v=H0r
where H 0 = 70 km s 1 Mpc 1 H 0 = 70 km s 1 Mpc 1 H_(0)=70kms^(-1)Mpc^(-1)H_{0}=70 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}H0=70 km s1Mpc1 is Hubble's constant. 9 9 ^(9){ }^{9}9 The interpretation of this law is that the Universe is expanding as a function of time. Two important points follow:
  • Hubble's law makes it look like our own position in the Universe is special, because everything in the Universe is receding from us. Does this violate the Copernican principle? Not at all, because Hubble's law holds viewed from everywhere in the Universe. One way of thinking about this is that for two objects, currently (time t 0 t 0 t_(0)t_{0}t0 ) at positions r i r i vec(r)_(i)\vec{r}_{i}ri and r j r j vec(r)_(j)\vec{r}_{j}rj, have a separation given by the vector r i j = r i r j r i j = r i r j vec(r)_(ij)= vec(r)_(i)- vec(r)_(j)\vec{r}_{i j}=\vec{r}_{i}-\vec{r}_{j}rij=rirj (and this description subtracts out the choice of coordinate origin that you have when you write down a particular r i r i vec(r)_(i)\vec{r}_{i}ri or r j r j vec(r)_(j)\vec{r}_{j}rj ). We now suppose that the Universe expands in such a way that, at some later time t t ttt, we have
(15.4) r ( t ) = a ( t ) r ( t 0 ) , (15.4) r ( t ) = a ( t ) r t 0 , {:(15.4) vec(r)(t)=a(t) vec(r)(t_(0))",":}\begin{equation*} \vec{r}(t)=a(t) \vec{r}\left(t_{0}\right), \tag{15.4} \end{equation*}(15.4)r(t)=a(t)r(t0),
where a ( t ) a ( t ) a(t)a(t)a(t) is the scale factor of the Universe. The function a ( t ) a ( t ) a(t)a(t)a(t) follows the expansion of the Universe and we can define it so that a ( t 0 ) = 1 a t 0 = 1 a(t_(0))=1a\left(t_{0}\right)=1a(t0)=1 (i.e. the Universe is not scaled up or down at the current time t 0 t 0 t_(0)t_{0}t0 ). Then the separation of the two objects will be given by
(15.5) r i j ( t ) = a ( t ) r i j ( t 0 ) (15.5) r i j ( t ) = a ( t ) r i j t 0 {:(15.5) vec(r)_(ij)(t)=a(t) vec(r)_(ij)(t_(0)):}\begin{equation*} \vec{r}_{i j}(t)=a(t) \vec{r}_{i j}\left(t_{0}\right) \tag{15.5} \end{equation*}(15.5)rij(t)=a(t)rij(t0)
and so the separation of any two objects increases according to the scale factor a ( t ) a ( t ) a(t)a(t)a(t), irrespective of any choice of origin. Moreover, this equation can be used to work out the relative velocity between these two objects as
(15.6) v i j ( t ) = r ˙ i j ( t ) = a ˙ ( t ) r i j ( t 0 ) = a ˙ ( t ) a ( t ) r i j ( t ) (15.6) v i j ( t ) = r ˙ i j ( t ) = a ˙ ( t ) r i j t 0 = a ˙ ( t ) a ( t ) r i j ( t ) {:(15.6) vec(v)_(ij)(t)= vec(r)^(˙)_(ij)(t)=a^(˙)(t) vec(r)_(ij)(t_(0))=((a^(˙))(t))/(a(t)) vec(r)_(ij)(t):}\begin{equation*} \vec{v}_{i j}(t)=\dot{\vec{r}}_{i j}(t)=\dot{a}(t) \vec{r}_{i j}\left(t_{0}\right)=\frac{\dot{a}(t)}{a(t)} \vec{r}_{i j}(t) \tag{15.6} \end{equation*}(15.6)vij(t)=r˙ij(t)=a˙(t)rij(t0)=a˙(t)a(t)rij(t)
and so we deduce that the Hubble parameter is
(15.7) H ( t ) = a ˙ ( t ) a ( t ) (15.7) H ( t ) = a ˙ ( t ) a ( t ) {:(15.7)H(t)=((a^(˙))(t))/(a(t)):}\begin{equation*} H(t)=\frac{\dot{a}(t)}{a(t)} \tag{15.7} \end{equation*}(15.7)H(t)=a˙(t)a(t)
We will call r ( t ) { = [ x i ( t ) , y i ( t ) , z i ( t ) ] } r ( t ) = x i ( t ) , y i ( t ) , z i ( t ) vec(r)(t){=[x^(i)(t),y^(i)(t),z^(i)(t)]}\vec{r}(t)\left\{=\left[x^{i}(t), y^{i}(t), z^{i}(t)\right]\right\}r(t){=[xi(t),yi(t),zi(t)]} the real, physical, coordinates and r ( t 0 ) r t 0 vec(r)(t_(0))\vec{r}\left(t_{0}\right)r(t0) the comoving coordinates. 10 10 ^(10){ }^{10}10 Galaxies generally remain fixed in place when measured in the comoving coordinates, but becomes increasingly separated from one another in the physical coordinates, as a ( t ) a ( t ) a(t)a(t)a(t) increases.
  • We use the term Hubble flow to denote the outward motion of objects due to the expansion of the Universe, as expressed by the scale factor a ( t ) a ( t ) a(t)a(t)a(t). There can be other so-called proper motions superimposed on top of the Hubble flow. For example, some galaxy may be gravitationally bound to some others in a cluster. Its velocity, resulting from forces within that system of galaxies, will be superimposed upon the velocity resulting from the Hubble flow. Another example is that the cosmic microwave background (CMB) radiation, observable from the Earth, is remarkably uniform across the sky but has a small dipole anisotropy superimposed on it. This results from the motion of the Earth with respect to the reference frame of the CMB, 11 11 ^(11){ }^{11}11 meaning that the CMB viewed from Earth is very slightly redshifted or blueshifted depending on which direction you look.
Note that the expansion of the Universe, governed by a ( t ) a ( t ) a(t)a(t)a(t), doesn't affect the Solar System. 12 12 ^(12){ }^{12}12 The physics of things on smaller length scales, such as the orbits of gravitationally bound objects in the Solar System, can be treated by general relativity (see Part IV of this book), but are dominated by the effects of the local interactions between the objects; the weak curvature of the Universe plays no role. Thus, in this part of the book (Part III) we are keeping our eyes on the big picture and ignoring the tiny details that only affect processes on minuscule length scales. 13 13 ^(13){ }^{13}13
10 A 10 A ^(10)A{ }^{10} \mathrm{~A}10 A simple working definition of comoving coordinates is that a particle at rest in a comoving system continues to be at rest as a function of coordinate time t t ttt. In this case, we have r ˙ i j ( t 0 ) = 0 r ˙ i j t 0 = 0 vec(r)^(˙)_(ij)(t_(0))=0\dot{\vec{r}}_{i j}\left(t_{0}\right)=0r˙ij(t0)=0
11 11 ^(11){ }^{11}11 Sometimes described as the echo of the Big Bang, the CMB is electromagnetic radiation resulting from recombination processes that occurred when the first atoms were formed, during a relatively early epoch of the Universe. Although special relativity teaches us that there's no privileged frame of reference, the CMB frame is an interesting case of a rather physically unique frame! Of course, the laws of physics are the same in that frame as in any other inertial frame. The small anisotropy in the CMB that we measure results from the motion of the Earth around the Sun (a time-dependent term with a period of one year) and the motion of the Solar System with respect to tion of the Solar System with respect to
the CMB, which includes the motion of the CMB, which includes the motion of
the Local Group of galaxies, including our own Milky Way, towards the centre of the constellation Hydra.
12 Or 12 Or ^(12)Or{ }^{12} \mathrm{Or}12Or, for that matter, the expansion of your waistline. That cannot be blamed on Edwin Hubble.
13 13 ^(13){ }^{13}13 By minuscule, we here mean less than 100 Mpc !
14 14 ^(14){ }^{14}14 This idea assumes that the Universe is evolving monotonically towards some final state and that there are characteristic motions that can be used as a standard of time. This give us access to some scalar S S SSS that decreases monotonically as the Universe evolves forward in time. An example might be the black-body radiation temperature T T TTT of the Universe: since the Universe is assumed to be cooling monotonically, T T TTT should steadily decrease as a function of the age of the Universe. The time t t ttt assigned to an event is then defined to be a function t ( T ) t ( T ) t(T)t(T)t(T) where T T TTT is evaluated when and where the event takes place. Since the scalar S S SSS used to characterize cosmic standard time t t ttt is a function of t t ttt only, this implies that for an event that is assigned time t t ttt in cosmic standard, and t t t^(')t^{\prime}t in the other coordinate system, we have S ( t ) = S ( t ) S ( t ) = S t S(t)=S^(')(t^('))S(t)=S^{\prime}\left(t^{\prime}\right)S(t)=S(t). However, since S S SSS is a scalar, we have S = S S = S S^(')=SS^{\prime}=SS=S and so S ( t ) = S ( t ) S ( t ) = S t S(t)=S(t^('))S(t)=S\left(t^{\prime}\right)S(t)=S(t). We see then that t = t t = t t=t^(')t=t^{\prime}t=t, and that all equivalent coordinate sys tems must use cosmic standard time.
Fig. 15.1 Spacelike hypersurfaces foliate the whole of spacetime. Homogeneity is defined in terms of spacelike hypersurfaces, so each spatial coordinate p p ppp on a given hypersurface is equivalent to every other one on that surface, e.g. point q q qqq. Points p p ppp and q q qqq on the same hypersurface are related by an isometry of spacetime which keeps time t t ttt fixed. An isometry is a distance-preserving transformation between parts of a spacetime.

15.3 Cosmic time

We shall now set up a coordinate system with which to investigate cosmology. We start with the notion of a cosmic time coordinate t t ttt that could reasonably be used by sensible observational astronomers. 14 14 ^(14){ }^{14}14 We can then define a cosmic standard coordinate system in terms of a stack of spacelike hypersurfaces at various fixed values of t t ttt. A hypersurface is a slice of spacetime formed by our keeping one of the four spacetime coordinates fixed so that the remaining coordinates describe a three-dimensional space. A spacelike hypersurface for a particular time t t ttt is then described by three spatial coordinates ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z), so that one spacelike hypersurface passes through each and every point in 3 -space. The spacelike hypersurfaces are said to foliate all of spacetime, meaning that spacetime is carved up into a stack of non-intersecting spacelike hypersurfaces, as shown in Fig. 15.1. The cosmological principle (homogeneity and isotropy) holds in each of these spacelike hypersurfaces, and so at each moment in time (the coordinate which labels which slice you are considering) the Universe has the same density, pressure and curvature everywhere. This means that in each hypersurface the Universe looks the same everywhere (satisfying our assumption of homogeneity) and in every direction (satisfying our assumption of isotropy, see Fig. 15.2), at least on length scales greater than about 100 Mpc (as discussed before). Of course, if you zoom along in a spacecraft travelling at some large speed with respect to this cosmic standard coordinate system (in other words, if you travel with some proper motion with respect to the Hubble flow) then the Universe will not look isotropic and homogeneous. Thus, when doing cosmology we will restrict our attention to the cosmic standard coordinate system where the cosmological principle works.
We met the idea of comoving coordinates in the last section. We shall see that a good choice of cosmic coordinate system is a comoving one. An observer who is at rest with respect to comoving coordinates (and who therefore 'goes with the flow' of the cosmic evolution, which you could describe colloquially as 'going with the Hubble flow') is known as a comoving observer. The comoving observers have world lines that coincide with the elements of the cosmological fluid, that pierce the spacelike hypersurfaces, as shown in Fig. 15.3.

Example 15.2

Putting the notions of homogeneity and isotropy together, it seems reasonable that spacelike hypersurfaces must be orthogonal to the world lines of the cosmological fluid. If they weren't we would be able to identify a preferred direction, contradicting the assumption of isotropy. This construction allows us geometrically to assert that the time coordinate in the comoving coordinates has mathematical properties:
(15.8) e 0 e i = g 0 i = 0 (15.8) e 0 e i = g 0 i = 0 {:(15.8)e_(0)*e_(i)=g_(0i)=0:}\begin{equation*} \boldsymbol{e}_{0} \cdot \boldsymbol{e}_{i}=g_{0 i}=0 \tag{15.8} \end{equation*}(15.8)e0ei=g0i=0
(i.e. the timelike vector is perpendicular to the spacelike ones);
(15.9) e 0 = u (15.9) e 0 = u {:(15.9)e_(0)=u:}\begin{equation*} e_{0}=u \tag{15.9} \end{equation*}(15.9)e0=u
(i.e. the comoving observer has a velocity that coincides with the timelike vector)
e 0 e 0 = g 00 = 1 e 0 e 0 = g 00 = 1 e_(0)*e_(0)=g_(00)=-1\boldsymbol{e}_{0} \cdot \boldsymbol{e}_{0}=g_{00}=-1e0e0=g00=1
(15.10)
(i.e. the timelike vector is normalized).
If a metric yields Γ i 00 = 0 Γ i 00 = 0 Gamma^(i)_(00)=0\Gamma^{i}{ }_{00}=0Γi00=0 then, by the geodesic equation ( x ¨ μ + Γ μ α β x ˙ α x ˙ β = 0 x ¨ μ + Γ μ α β x ˙ α x ˙ β = 0 x^(¨)^(mu)+Gamma^(mu)_(alpha beta)x^(˙)^(alpha)x^(˙)^(beta)=0\ddot{x}^{\mu}+\Gamma^{\mu}{ }_{\alpha \beta} \dot{x}^{\alpha} \dot{x}^{\beta}=0x¨μ+Γμαβx˙αx˙β=0 ), a particle at rest in this coordinate system stays at rest. This is then a comoving coordinate system. This connection coefficient will vanish if the derivatives g 00 , i g 00 , i g_(00,i)g_{00, i}g00,i and g 0 i , 0 g 0 i , 0 g_(0i,0)g_{0 i, 0}g0i,0 vanish, giving us a quick way of identifying comoving coordinates. So in the present case, we can conclude that the conditions for the coordinates to be comoving ( g 0 i , 0 = g 00 , i = 0 g 0 i , 0 = g 00 , i = 0 g_(0i,0)=g_(00,i)=0g_{0 i, 0}=g_{00, i}=0g0i,0=g00,i=0 ) are met.
Comoving observers have a special place in our cosmological theories as they form an important part of the underlying geometry, as enshrined in Weyl's postulate, which states that: 15 15 ^(15){ }^{15}15
In cosmic spacetime, there exists a set of privileged fundamental observers whose world lines form a smooth bundle of timelike geodesics. These geodesics never meet (apart from at an initial, past singularity or a final, future singularity).
This state of affairs is shown in Fig. 15.3. In comoving coordinates, each galaxy is at rest (ignoring proper motions) and each hypersurface is assigned a universal cosmic time coordinate. This time coincides with the proper time of the privileged observer at that point.
Finally, we consider the consequences of the principle of cosmology on the form of the fundamental field in general relativity: the metric. This will be expressed in our comoving coordinates x μ = ( t , x i ) x μ = t , x i x^(mu)=(t,x^(i))x^{\mu}=\left(t, x^{i}\right)xμ=(t,xi). To retain homogeneity and isotropy, the metric line element d σ d σ dsigma\mathrm{d} \sigmadσ of each of the three-dimensional, spacelike hypersurfaces must take the form d σ 2 = γ i j d x i d x j d σ 2 = γ i j d x i d x j dsigma^(2)=gamma_(ij)dx^(i)dx^(j)\mathrm{d} \sigma^{2}=\gamma_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j}dσ2=γij dxi dxj, where γ i j γ i j gamma_(ij)\gamma_{i j}γij is a function of spatial coordinates x 1 , x 2 x 1 , x 2 x^(1),x^(2)x^{1}, x^{2}x1,x2 and x 3 x 3 x^(3)x^{3}x3 and encodes information about the curvature of the space. 16 16 ^(16){ }^{16}16 Also including the time, the metric line element of spacetime in the general comoving system is given by
(15.11) d s 2 = d t 2 + a ( t ) 2 γ i j d x i d x j (15.11) d s 2 = d t 2 + a ( t ) 2 γ i j d x i d x j {:(15.11)ds^(2)=-dt^(2)+a(t)^(2)gamma_(ij)dx^(i)dx^(j):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2} \gamma_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j} \tag{15.11} \end{equation*}(15.11)ds2=dt2+a(t)2γij dxi dxj
Here, we have included a ( t ) a ( t ) a(t)a(t)a(t), the time-dependent scaling factor, that describes the expansion of space as a function of time. 17 17 ^(17){ }^{17}17 This kinematic approach to the metric was pioneered by Robertson and Walker in the 1930 s 18 1930 s 18 1930s^(18)1930 \mathrm{~s}^{18}1930 s18 and we shall return to its details in the next chapter. To obtain the form of a ( t ) a ( t ) a(t)a(t)a(t) we must make assumptions about the content of the Universe. This is the subject to which we now turn. It's time to make some model universes! Remember that our goal is to obtain a solution to the Einstein field equation, which could be that for a given geometry or for a given distribution of mass and energy.

15.4 Universe 0: an empty universe

The first and simplest model describes an empty universe, containing no matter and no energy, so that T μ ν = 0 T μ ν = 0 T^(mu nu)=0T^{\mu \nu}=0Tμν=0. We therefore have an Einstein equation
(15.12) G μ ν = R μ ν 1 2 g μ ν R = 0 (15.12) G μ ν = R μ ν 1 2 g μ ν R = 0 {:(15.12)G_(mu nu)=R_(mu nu)-(1)/(2)g_(mu nu)R=0:}\begin{equation*} G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R=0 \tag{15.12} \end{equation*}(15.12)Gμν=Rμν12gμνR=0
Fig. 15.2 World lines run perpendicular to the spacelike hypersurfaces and isotropy means that, at some point P P P\mathcal{P}P, the Universe looks the same along any two spacelike vectors e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 and e 2 e 2 e_(2)\boldsymbol{e}_{2}e2. We've drawn them to be roughly perpendicular to each other, but they needn't be. However, both have to be spacelike and so both must be perpendicular to e 0 e 0 e_(0)e_{0}e0, the timelike vector that is tangent to the world line. There exists an isometry in spacetime which leaves P P P\mathcal{P}P fixed but which rotates e 1 e 1 e_(1)e_{1}e1 into e 2 e 2 e_(2)e_{2}e2.
15 15 ^(15){ }^{15}15 Hermann Weyl (1885-1955).
Smoothness' in this statement guarantees that our classical field theory can be used to describe the physics. The word 'bundle' here has a technical meaning examined in Appendix C, but meaning examined in Appendix c, but of the word, as 'a collection of things of the word, as 'a collection of things
fastened together for convenient hanfastene
dling'.
Fig. 15.3 Timelike geodesics of cosmic observers. These are orthogonal to all of the spacelike hypersurfaces.
16 16 ^(16){ }^{16}16 For flat space we have γ 11 = γ 22 = γ 11 = γ 22 = gamma_(11)=gamma_(22)=\gamma_{11}=\gamma_{22}=γ11=γ22= γ 33 = 1 γ 33 = 1 gamma_(33)=1\gamma_{33}=1γ33=1 and other components zero.
17 17 ^(17){ }^{17}17 Check: g 0 i , 0 = g 00 , i = 0 g 0 i , 0 = g 00 , i = 0 g_(0i,0)=g_(00,i)=0g_{0 i, 0}=g_{00, i}=0g0i,0=g00,i=0, so the system is confirmed as comoving.
18 18 ^(18){ }^{18}18 Howard P. 'Bob' Robertson (19031961); Arthur G. Walker (1909-2001).
19 19 ^(19){ }^{19}19 However, the condition R μ ν = 0 R μ ν = 0 R_(mu nu)=0R_{\mu \nu}=0Rμν=0 on its own does not mean that the components of the Riemann tensor must also vanish (giving a flat universe, as they do in this model); rather that the positive curvature parts cancel out the negative curvature parts when the Ricc tensor is formed. Physically, the Ricci tensor describes the tendency of gravitational fields to make volumes of mass contract. This is described in more de tail in Chapter 35. The photograph of Einstein writing the equation R i k = R i k = R_(ik)=R_{i k}=Rik= 0 ?' on a blackboard has become iconic and might, after E = m E = m E=mE=mE=m (or E = m c 2 E = m c 2 E=mc^(2)E=m c^{2}E=mc2 in units that Einstein didn't generally use), have a claim to being his second most-famous equation.
20 20 ^(20){ }^{20}20 Expressed in spherical polars, this metric is
d s 2 = d t 2 + d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = d t 2 + d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-dt^(2)+dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))\mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)ds2=dt2+dr2+r2( dθ2+sin2θ dϕ2).
This can be rewritten using light-cone coordinates u = t r u = t r u=t-ru=t-ru=tr and v = t + r v = t + r v=t+rv=t+rv=t+r as
d s 2 = d u d v (15.14) + 1 4 ( v u ) 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = d u d v (15.14) + 1 4 ( v u ) 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-dudv],[(15.14)+(1)/(4)(v-u)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{align*} & \mathrm{d} s^{2}=-\mathrm{d} u \mathrm{~d} v \\ & +\frac{1}{4}(v-u)^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{15.14} \end{align*}ds2=du dv(15.14)+14(vu)2( dθ2+sin2θ dϕ2)
We shall make use of each of these forms in the coming chapters.
21 21 ^(21){ }^{21}21 Willem de Sitter (1872-1934). We discuss a neat geometrical description of the this model's spacetime in the exercises of Chapter 18.
22 22 ^(22){ }^{22}22 Fixing the time at t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0 results in a spacelike hypersurface with d σ 2 = d σ 2 = dsigma^(2)=\mathrm{d} \sigma^{2}=dσ2= a ( t 0 ) ( d x 2 + d y 2 + d z 2 ) a t 0 d x 2 + d y 2 + d z 2 a(t_(0))(dx^(2)+dy^(2)+dz^(2))a\left(t_{0}\right)\left(\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2}\right)a(t0)(dx2+dy2+dz2), which is the homogeneous, isotropic metric we had in Universe 0, scaled by a factor a ( t 0 ) a t 0 a(t_(0))a\left(t_{0}\right)a(t0). Notice how the expansion factor a ( t ) a ( t ) a(t)a(t)a(t) affects the behaviour of light cones, which obey d t / d r = ± a ( t ) d t / d r = ± a ( t ) dt//dr=+-a(t)\mathrm{d} t / \mathrm{d} r= \pm a(t)dt/dr=±a(t). For a monotonically increasing a ( t ) a ( t ) a(t)a(t)a(t) this causes light cones to become narrower with time This effect is examined in Chapter 25.
Since R = g μ ν R μ ν R = g μ ν R μ ν R=g_(mu nu)R^(mu nu)R=g_{\mu \nu} R^{\mu \nu}R=gμνRμν, we see that R μ ν = 0 R μ ν = 0 R_(mu nu)=0R_{\mu \nu}=0Rμν=0 guarantees a solution to the Einstein equation. 19 19 ^(19){ }^{19}19
A simple solution to the Einstein equation that is consistent with an empty, isotropic universe is flat, Minkowski spacetime with metric 20 20 ^(20){ }^{20}20
(15.15) d s 2 = d t 2 + d x 2 + d y 2 + d z 2 (15.15) d s 2 = d t 2 + d x 2 + d y 2 + d z 2 {:(15.15)ds^(2)=-dt^(2)+dx^(2)+dy^(2)+dz^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2} \tag{15.15} \end{equation*}(15.15)ds2=dt2+dx2+dy2+dz2
This flat spacetime, which is written in comoving coordinates (i.e. g 00 , i = g 00 , i = g_(00,i)=g_{00, i}=g00,i= g 0 i , 0 = 0 g 0 i , 0 = 0 g_(0i,0)=0g_{0 i, 0}=0g0i,0=0 ), is consistent with eqn 15.11, and has R μ ν α β = 0 R μ ν α β = 0 R_(mu nu alpha beta)=0R_{\mu \nu \alpha \beta}=0Rμναβ=0 and therefore R μ ν = 0 R μ ν = 0 R_(mu nu)=0R_{\mu \nu}=0Rμν=0. We conclude that flat space is a solution to this problem. Hypersurfaces of this spacetime are spatially flat with d σ 2 = d x 2 + d σ 2 = d x 2 + dsigma^(2)=dx^(2)+\mathrm{d} \sigma^{2}=\mathrm{d} x^{2}+dσ2=dx2+ d y 2 + d z 2 d y 2 + d z 2 dy^(2)+dz^(2)\mathrm{d} y^{2}+\mathrm{d} z^{2}dy2+dz2. Geodesics in this universe take the form of straight lines
(15.16) x μ ( τ ) = b μ τ + c μ (15.16) x μ ( τ ) = b μ τ + c μ {:(15.16)x^(mu)(tau)=b^(mu)tau+c^(mu):}\begin{equation*} x^{\mu}(\tau)=b^{\mu} \tau+c^{\mu} \tag{15.16} \end{equation*}(15.16)xμ(τ)=bμτ+cμ
with b μ b μ b^(mu)b^{\mu}bμ and c μ c μ c^(mu)c^{\mu}cμ constants and τ τ tau\tauτ the proper time parametrizing the geodesic.
So far so good, but this energy/matter-free model is rather unrealistic. We can go much further by investigating a more interesting scenario.

15.5 Universe 1: flat and expanding

We next discuss a universe where space is flat, but expands as a function of time with expansion factor a ( t ) a ( t ) a(t)a(t)a(t). We'll call this the de Sitter model. 21 21 ^(21){ }^{21}21 The model's metric line element can be written as
d s 2 = d t 2 + a ( t ) 2 ( d x 2 + d y 2 + d z 2 ) (15.17) = d t 2 + a ( t ) 2 [ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] d s 2 = d t 2 + a ( t ) 2 d x 2 + d y 2 + d z 2 (15.17) = d t 2 + a ( t ) 2 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-dt^(2)+a(t)^(2)((d)x^(2)+dy^(2)+dz^(2))],[(15.17)=-dt^(2)+a(t)^(2)[(d)r^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]]:}\begin{align*} \mathrm{d} s^{2} & =-\mathrm{d} t^{2}+a(t)^{2}\left(\mathrm{~d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2}\right) \\ & =-\mathrm{d} t^{2}+a(t)^{2}\left[\mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{15.17} \end{align*}ds2=dt2+a(t)2( dx2+dy2+dz2)(15.17)=dt2+a(t)2[ dr2+r2( dθ2+sin2θ dϕ2)]
where, in the second line we've employed spherical polar coordinates. This is another isotropic and homogeneous model of spacetime. 22 22 ^(22){ }^{22}22 The dimensionless factor a ( t ) a ( t ) a(t)a(t)a(t), if positive, causes space itself to expand, such that all of the coordinate points (or all of the galaxies) move away from each other. Although the spacelike hypersurfaces of this model are flat, spacetime is not (i.e. there will be some non-zero components of R R R\boldsymbol{R}R ). Therefore, our first task it to work out how the scaling factor a ( t ) a ( t ) a(t)a(t)a(t) affects the curvature of spacetime.

Example 15.3

We can use the metric field of the de Sitter model to compute the curvature of spacetime. The metric components are
(15.18) g 00 = 1 , g i i = a 2 , g 00 = 1 , g i i = 1 a 2 . (15.18) g 00 = 1 , g i i = a 2 , g 00 = 1 , g i i = 1 a 2 . {:(15.18){:[g_(00)=-1",",g_(ii)=a^(2)","],[g^(00)=-1",",g^(ii)=(1)/(a^(2)).]:}:}\begin{array}{ll} g_{00}=-1, & g_{i i}=a^{2}, \tag{15.18}\\ g^{00}=-1, & g^{i i}=\frac{1}{a^{2}} . \end{array}(15.18)g00=1,gii=a2,g00=1,gii=1a2.
The derivation of the connection coefficients and the components of the Riemann tensor will be covered in the Exercises. Here we'll get some practice using the orthonormal frame and the vielbein. The metric field yields up the components of the Riemann tensor in the orthonormal frame of
(15.19) R i ^ 0 ^ i ^ i ^ ^ = a ¨ a , R j ^ i ^ j ^ i ^ = ( a ˙ a ) 2 , (15.19) R i ^ 0 ^ i ^ i ^ ^ = a ¨ a , R j ^ i ^ j ^ i ^ = a ˙ a 2 , {:(15.19)R_( hat(i) hat(0) hat(i))^( hat(hat(i)))=((a^(¨)))/(a)","quadR_( hat(j) hat(i) hat(j))^( hat(i))=(((a^(˙)))/(a))^(2)",":}\begin{equation*} R_{\hat{i} \hat{0} \hat{i}}^{\hat{\hat{i}}}=\frac{\ddot{a}}{a}, \quad R_{\hat{j} \hat{i} \hat{j}}^{\hat{i}}=\left(\frac{\dot{a}}{a}\right)^{2}, \tag{15.19} \end{equation*}(15.19)Ri^0^i^i^^=a¨a,Rj^i^j^i^=(a˙a)2,
where a ˙ = d a ( t ) / d t a ˙ = d a ( t ) / d t a^(˙)=da(t)//dt\dot{a}=\mathrm{d} a(t) / \mathrm{d} ta˙=da(t)/dt and we (temporarily) do not sum over repeated indices.
where a ˙ = d a ( t ) / d t a ˙ = d a ( t ) / d t a^(˙)=da(t)//dt\dot{a}=\mathrm{d} a(t) / \mathrm{d} ta˙=da(t)/dt and we (temporarily) do not sum over repeated indices.
We can identify the vielbein 23 23 ^(23){ }^{23}23 allowing us to move between the orthonormal and coordinate frames. The Ricci tensor's 00th component in the orthonormal frame is
(15.21) R 0 0 ^ = i ^ = 1 3 R 0 ^ i ^ 0 ^ i ^ (15.21) R 0 0 ^ = i ^ = 1 3 R 0 ^ i ^ 0 ^ i ^ {:(15.21)R_(0 hat(0))=sum_( hat(i)=1)^(3)R_( hat(0) hat(i) hat(0))^( hat(i)):}\begin{equation*} R_{0 \hat{0}}=\sum_{\hat{i}=1}^{3} R_{\hat{0} \hat{i} \hat{0}}^{\hat{i}} \tag{15.21} \end{equation*}(15.21)R00^=i^=13R0^i^0^i^
where we write the sum explicitly. Using the idea that components are raised and lowered in the orthonormal frame using the Minkowski tensor η η eta\boldsymbol{\eta}η, we have from eqn 15.19 that
(15.22) a ¨ a = R 0 ^ i ^ 0 ^ i ^ ( using η 00 = 1 ) = R i ^ i ^ i ^ 0 ^ ( two signs from s = R 0 ^ i ^ 0 ^ ( using η i i = 1 ) . (15.22) a ¨ a = R 0 ^ i ^ 0 ^ i ^  using  η 00 = 1 = R i ^ i ^ i ^ 0 ^ (  two   signs from  s = R 0 ^ i ^ 0 ^  using  η i i = 1 . {:(15.22){:[((a^(¨)))/(a),=-R_( hat(0) hat(i) hat(0) hat(i)),,],[,(" using "eta_(00)=-1)],[,=-R_( hat(i) hat(i) hat(i) hat(0)),,(" two "-" signs from "s],[,=-R_( hat(0) hat(i) hat(0)),,(" using "eta^(ii)=1).]:}:}\begin{array}{rlrl} \frac{\ddot{a}}{a} & =-R_{\hat{0} \hat{i} \hat{0} \hat{i}} & & \tag{15.22}\\ & \left(\text { using } \eta_{00}=-1\right) \\ & =-R_{\hat{i} \hat{i} \hat{i} \hat{0}} & & (\text { two }- \text { signs from } s \\ & =-R_{\hat{0} \hat{i} \hat{0}} & & \left(\text { using } \eta^{i i}=1\right) . \end{array}(15.22)a¨a=R0^i^0^i^( using η00=1)=Ri^i^i^0^( two  signs from s=R0^i^0^( using ηii=1).
we find
(15.23) R 0 ^ 0 ^ = 3 a ¨ a (15.23) R 0 ^ 0 ^ = 3 a ¨ a {:(15.23)R_( hat(0) hat(0))=-3((a^(¨)))/(a):}\begin{equation*} R_{\hat{0} \hat{0}}=-3 \frac{\ddot{a}}{a} \tag{15.23} \end{equation*}(15.23)R0^0^=3a¨a
It will be useful later to have the components in the coordinate frame, and so we transform out of the orthonormal frame using the vielbein
(15.24) R 00 = ( e 0 ) 0 ^ ( e 0 ) 0 ^ R 0 0 ^ = 3 a ¨ a (15.24) R 00 = e 0 0 ^ e 0 0 ^ R 0 0 ^ = 3 a ¨ a {:(15.24)R_(00)=(e_(0))^( hat(0))(e_(0))^( hat(0))R_(0 hat(0))=-3((a^(¨)))/(a):}\begin{equation*} R_{00}=\left(e_{0}\right)^{\hat{0}}\left(e_{0}\right)^{\hat{0}} R_{0 \hat{0}}=-3 \frac{\ddot{a}}{a} \tag{15.24} \end{equation*}(15.24)R00=(e0)0^(e0)0^R00^=3a¨a
Note that owing to the symmetries of R R R\boldsymbol{R}R we have, as always, that R μ μ μ μ = 0 R μ μ μ μ = 0 R^(mu)_(mu mu mu)=0R^{\mu}{ }_{\mu \mu \mu}=0Rμμμμ=0. As a result, to avoid a contribution from the case when i = j i = j i=ji=ji=j we only have two contributions from the spatial sum in the expression
R i ^ i ^ = R i ^ i ^ ^ i ^ 0 ^ + j = 1 2 R i ^ j ^ i ^ j ^ (15.25) = a ¨ a + 2 ( a ˙ a ) 2 , R i ^ i ^ = R i ^ i ^ ^ i ^ 0 ^ + j = 1 2 R i ^ j ^ i ^ j ^ (15.25) = a ¨ a + 2 a ˙ a 2 , {:[R_( hat(i) hat(i))=R_( hat(i) hat(hat(i)) hat(i))^( hat(0))+sum_(j=1)^(2)R_( hat(i) hat(j) hat(i))^( hat(j))],[(15.25)=((a^(¨)))/(a)+2(((a^(˙)))/(a))^(2)","]:}\begin{align*} R_{\hat{i} \hat{i}} & =R_{\hat{i} \hat{\hat{i}} \hat{i}}^{\hat{0}}+\sum_{j=1}^{2} R_{\hat{i} \hat{j} \hat{i}}^{\hat{j}} \\ & =\frac{\ddot{a}}{a}+2\left(\frac{\dot{a}}{a}\right)^{2}, \tag{15.25} \end{align*}Ri^i^=Ri^i^^i^0^+j=12Ri^j^i^j^(15.25)=a¨a+2(a˙a)2,
and so, transforming into the coordinate frame,
R i i = ( e i ) i ^ ( e i ) i ^ R i ^ i ^ = a 2 R i ^ i ^ R i i = e i i ^ e i i ^ R i ^ i ^ = a 2 R i ^ i ^ R_(ii)=(e_(i))^( hat(i))(e_(i))^( hat(i))R_( hat(i) hat(i))=a^(2)R_( hat(i) hat(i))R_{i i}=\left(\boldsymbol{e}_{i}\right)^{\hat{i}}\left(\boldsymbol{e}_{i}\right)^{\hat{i}} R_{\hat{i} \hat{i}}=a^{2} R_{\hat{i} \hat{i}}Rii=(ei)i^(ei)i^Ri^i^=a2Ri^i^
(15.26) = a ¨ a + 2 a ˙ 2 (15.26) = a ¨ a + 2 a ˙ 2 {:(15.26)=a^(¨)a+2a^(˙)^(2):}\begin{equation*} =\ddot{a} a+2 \dot{a}^{2} \tag{15.26} \end{equation*}(15.26)=a¨a+2a˙2
Working in any frame, we must obtain the same Ricci scalar. In the coordinate frame, we find
R = g μ ν R μ ν = 3 a ¨ a + 3 a 2 ( 2 a ˙ 2 + a a ¨ ) (15.27) = 6 ( a ¨ a + a ˙ 2 a 2 ) R = g μ ν R μ ν = 3 a ¨ a + 3 a 2 2 a ˙ 2 + a a ¨ (15.27) = 6 a ¨ a + a ˙ 2 a 2 {:[R=g^(mu nu)R_(mu nu)=3((a^(¨)))/(a)+(3)/(a^(2))*(2a^(˙)^(2)+a(a^(¨)))],[(15.27)=6(((a^(¨)))/(a)+(a^(˙)^(2))/(a^(2)))]:}\begin{align*} R=g^{\mu \nu} R_{\mu \nu} & =3 \frac{\ddot{a}}{a}+\frac{3}{a^{2}} \cdot\left(2 \dot{a}^{2}+a \ddot{a}\right) \\ & =6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right) \tag{15.27} \end{align*}R=gμνRμν=3a¨a+3a2(2a˙2+aa¨)(15.27)=6(a¨a+a˙2a2)
What matter distribution could be responsible for this spacetime metric and provide the right-hand side of the Einstein equation? It turns out that pure vacuum energy is consistent with this metric as we shall now see. If we allow the presence of vacuum energy but not matter, then our task is to solve an Einstein equation
(15.28) R μ ν 1 2 g μ ν R = Λ g μ ν (15.28) R μ ν 1 2 g μ ν R = Λ g μ ν {:(15.28)R_(mu nu)-(1)/(2)g_(mu nu)R=-Lambdag_(mu nu):}\begin{equation*} R_{\mu \nu}-\frac{1}{2} g_{\mu \nu} R=-\Lambda g_{\mu \nu} \tag{15.28} \end{equation*}(15.28)Rμν12gμνR=Λgμν
with the Ricci tensor specified above.
23 23 ^(23){ }^{23}23 Recall that the orthonormal frame we choose is the one in which the observer's velocity vector u u u\boldsymbol{u}u only has a timelike component u μ = ( u 0 , 0 , 0 , 0 ) u μ = u 0 , 0 , 0 , 0 u^(mu)=(u^(0),0,0,0)u^{\mu}=\left(u^{0}, 0,0,0\right)uμ=(u0,0,0,0). Since we need u u = g μ ν u μ u ν = 1 u u = g μ ν u μ u ν = 1 u*u=g_(mu nu)u^(mu)u^(nu)=-1\boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} u^{\mu} u^{\nu}=-1uu=gμνuμuν=1, we must have u 0 = 1 u 0 = 1 u^(0)=1u^{0}=1u0=1. We then set e 0 ^ = u e 0 ^ = u e_( hat(0))=u\boldsymbol{e}_{\hat{0}}=\boldsymbol{u}e0^=u and e i ^ = 1 a e i e i ^ = 1 a e i e_( hat(i))=(1)/(a)e_(i)\boldsymbol{e}_{\hat{i}}=\frac{1}{a} \boldsymbol{e}_{i}ei^=1aei to give us orthonormal basis vectors. The vielbein components are
(15.20) ( e 0 ) 0 ^ = 1 , ( e i ) i ^ = a , ( e 0 ^ ) 0 = 1 , ( e i ^ ) i = 1 a . (15.20) e 0 0 ^ = 1 , e i i ^ = a , e 0 ^ 0 = 1 , e i ^ i = 1 a . {:(15.20){:[(e_(0))^( hat(0))=1",",(e_(i))^( hat(i))=a","],[(e_( hat(0)))^(0)=1",",(e_( hat(i)))^(i)=(1)/(a).]:}:}\begin{array}{ll} \left(\boldsymbol{e}_{0}\right)^{\hat{0}}=1, & \left(\boldsymbol{e}_{i}\right)^{\hat{i}}=a, \tag{15.20}\\ \left(\boldsymbol{e}_{\hat{0}}\right)^{0}=1, & \left(\boldsymbol{e}_{\hat{i}}\right)^{i}=\frac{1}{a} . \end{array}(15.20)(e0)0^=1,(ei)i^=a,(e0^)0=1,(ei^)i=1a.
Fig. 15.4 The function a ( t ) a ( t ) a(t)a(t)a(t) for Universe 1: the de Sitter model.
Fig. 15.5 Universe 1 is spatially flat with a scaling factor a ( t ) a ( t ) a(t)a(t)a(t) that increases exponentially as a function of time.
Example 15.4
Consider the 00 th component R 00 1 2 g 00 R = Λ g 00 R 00 1 2 g 00 R = Λ g 00 R_(00)-(1)/(2)g_(00)R=-Lambdag_(00)R_{00}-\frac{1}{2} g_{00} R=-\Lambda g_{00}R0012g00R=Λg00, which becomes, on substitution,
(15.29) 3 a ¨ a g 00 2 × 6 ( a ¨ a + a ˙ 2 a 2 ) = Λ g 00 . (15.29) 3 a ¨ a g 00 2 × 6 a ¨ a + a ˙ 2 a 2 = Λ g 00 . {:(15.29)-(3(a^(¨)))/(a)-(g_(00))/(2)xx6(((a^(¨)))/(a)+(a^(˙)^(2))/(a^(2)))=-Lambdag_(00).:}\begin{equation*} -\frac{3 \ddot{a}}{a}-\frac{g_{00}}{2} \times 6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right)=-\Lambda g_{00} . \tag{15.29} \end{equation*}(15.29)3a¨ag002×6(a¨a+a˙2a2)=Λg00.
This simplifies to
(15.30) 3 a ˙ 2 a 2 = Λ (15.30) 3 a ˙ 2 a 2 = Λ {:(15.30)3(a^(˙)^(2))/(a^(2))=Lambda:}\begin{equation*} 3 \frac{\dot{a}^{2}}{a^{2}}=\Lambda \tag{15.30} \end{equation*}(15.30)3a˙2a2=Λ
We can solve this straightforward first-order differential equation to find an equation for the scaling factor. The solution is
(15.31) a ( t ) = e H t (15.31) a ( t ) = e H t {:(15.31)a(t)=e^(Ht):}\begin{equation*} a(t)=\mathrm{e}^{H t} \tag{15.31} \end{equation*}(15.31)a(t)=eHt
with H 2 = Λ / 3 H 2 = Λ / 3 H^(2)=Lambda//3H^{2}=\Lambda / 3H2=Λ/3. As can be checked, this also solves the other components of the Einstein equation. The behaviour of a ( t ) a ( t ) a(t)a(t)a(t) is shown in Fig. 15.4. The increasing value reflects an accelerating expansion of this universe.
This is our first kinematic model of a universe. Granted, there's no matter in it yet (just energy), but we have found that adding uniform cosmological-constant energy density causes the universe to expand at an accelerating rate. The expansion constant H H HHH is an example of a Hubble parameter. The metric-field line element describing the geometry of this universe is, on substituting for a ( t ) a ( t ) a(t)a(t)a(t) and using polar coordinates,
(15.32) d s 2 = d t + e 2 H t [ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (15.32) d s 2 = d t + e 2 H t d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(15.32)ds^(2)=-dt+e^(2Ht)[(d)r^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t+\mathrm{e}^{2 H t}\left[\mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{15.32} \end{equation*}(15.32)ds2=dt+e2Ht[ dr2+r2( dθ2+sin2θ dϕ2)]
with H = ( Λ / 3 ) 1 2 H = ( Λ / 3 ) 1 2 H=(Lambda//3)^((1)/(2))H=(\Lambda / 3)^{\frac{1}{2}}H=(Λ/3)12. The evolution of Universe 1 is shown in Fig. 15.5, which shows the spatially flat universe expanding as a function of time.
Looked at another way, the (positive) vacuum-energy density described by Λ Λ Lambda\LambdaΛ has the curious property of causing an extra expansion to the universe. We can treat the vacuum energy as part of an energymomentum tensor, just as we did in Chapter 13 where we found that T μ ν vac = Λ g μ ν / 8 π T μ ν vac = Λ g μ ν / 8 π T_(mu nu)^(vac)=-Lambdag_(mu nu)//8piT_{\mu \nu}^{\mathrm{vac}}=-\Lambda g_{\mu \nu} / 8 \piTμνvac=Λgμν/8π. Comparing this to the components T μ = ( ρ , p , p , p ) T μ = ( ρ , p , p , p ) T^(mu)=(rho,p,p,p)T^{\mu}=(\rho, p, p, p)Tμ=(ρ,p,p,p) describing a perfect fluid in flat spacetime, we find an effective pressure p = Λ / 8 π p = Λ / 8 π p=-Lambda//8pip=-\Lambda / 8 \pip=Λ/8π. We conclude that vacuum energy with positive Λ Λ Lambda\LambdaΛ gives rise to a negative pressure and so, if squeezed, must expand.
We shall return to this model in Chapter 18. In the next two chapters, we shall put in place the groundwork that enables us to treat a far wider class of model universes.

Chapter summary

  • The cosmological principle says that the Universe is spatially homogeneous and isotropic.
  • An empty universe (Universe 0 ) has flat spacetime described by a Minkowski metric field.
  • The de Sitter model (Universe 1) is spatially flat, but has a spacetime with a constant curvature. It expands exponentially as a result of the vacuum energy content of the universe.

Exercises

(15.1) Taking a typical interstellar distance in the Milky Way as 1 pc , and the approximate dimensions of the Milky Way as given in Example 15.1, estimate the number of stars in the Milky Way. (This is astrophysics, so you only need to do this to the nearest power of ten.)
(15.2) Consider the de Sitter model.
(a) Show that the connection coefficients are given by
(15.33) Γ i j 0 = a a ˙ δ i j , Γ 0 j i = Γ j 0 i = a ˙ a δ j i , (15.33) Γ i j 0 = a a ˙ δ i j , Γ 0 j i = Γ j 0 i = a ˙ a δ j i , {:(15.33)Gamma_(ij)^(0)=aa^(˙)delta_(ij)","quadGamma_(0j)^(i)=Gamma_(j0)^(i)=((a^(˙)))/(a)delta_(j)^(i)",":}\begin{equation*} \Gamma_{i j}^{0}=a \dot{a} \delta_{i j}, \quad \Gamma_{0 j}^{i}=\Gamma_{j 0}^{i}=\frac{\dot{a}}{a} \delta_{j}^{i}, \tag{15.33} \end{equation*}(15.33)Γij0=aa˙δij,Γ0ji=Γj0i=a˙aδji,
where a ˙ = d a / d t a ˙ = d a / d t a^(˙)=da//dt\dot{a}=\mathrm{d} a / \mathrm{d} ta˙=da/dt.
(b) Compute the components of the curvature tensor R R R\boldsymbol{R}R.
(15.3) Another way to think about the de Sitter model is that it is defined to be a model with a constant curvature of spacetime, where the Ricci scalar R R RRR is sufficient to determine the Riemann tensor. The condition for such a constant curvature in (3+1)dimensional spacetime is that
(15.34) R μ ν = 1 4 R g μ ν (15.34) R μ ν = 1 4 R g μ ν {:(15.34)R_(mu nu)=(1)/(4)Rg_(mu nu):}\begin{equation*} R_{\mu \nu}=\frac{1}{4} R g_{\mu \nu} \tag{15.34} \end{equation*}(15.34)Rμν=14Rgμν
(a) Compute the components of the Einstein tensor in terms of R R RRR and g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν.
(b) Using this result, show that a solution to the Einstein equation requires Λ = R / 4 Λ = R / 4 Lambda=R//4\Lambda=R / 4Λ=R/4.
(c) Check the consistency of this last result using the expression R = 6 ( a ¨ a + a ˙ 2 a 2 ) R = 6 a ¨ a + a ˙ 2 a 2 R=6(((a^(¨)))/(a)+(a^(˙)^(2))/(a^(2)))R=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right)R=6(a¨a+a˙2a2), that we determined previously.
(15.4) A two-dimensional spacetime of constant curvature can be found by considering the metric line element we examined in Exercise 9.3
(15.35) d s 2 = d u 2 + d v 2 + 2 d u d v cos θ ( u , v ) (15.35) d s 2 = d u 2 + d v 2 + 2 d u d v cos θ ( u , v ) {:(15.35)ds^(2)=du^(2)+dv^(2)+2dudv cos theta(u","v):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} u^{2}+\mathrm{d} v^{2}+2 \mathrm{~d} u \mathrm{~d} v \cos \theta(u, v) \tag{15.35} \end{equation*}(15.35)ds2=du2+dv2+2 du dvcosθ(u,v)
It is possible to write the line element of any twodimensional spacetime in this form.
(a) Using the connection coefficients computed in Exercise 9.3, show that we have (using comma notation for derivatives)
(15.36) R v u v u = θ , u v sin θ , R u u v u = cos θ sin θ θ , u v (15.36) R v u v u = θ , u v sin θ , R u u v u = cos θ sin θ θ , u v {:(15.36)R_(vuv)^(u)=-(theta,uv)/(sin theta)","quadR_(uuv)^(u)=-(cos theta)/(sin theta)theta_(,uv):}\begin{equation*} R_{v u v}^{u}=-\frac{\theta, u v}{\sin \theta}, \quad R_{u u v}^{u}=-\frac{\cos \theta}{\sin \theta} \theta_{, u v} \tag{15.36} \end{equation*}(15.36)Rvuvu=θ,uvsinθ,Ruuvu=cosθsinθθ,uv
and that, consequently,
(15.37) R u u = θ , u v sin θ , R u v = cos θ sin θ θ , u v (15.37) R u u = θ , u v sin θ , R u v = cos θ sin θ θ , u v {:(15.37)R_(uu)=-(theta,uv)/(sin theta)","quadR_(uv)=-(cos theta)/(sin theta)theta_(,uv):}\begin{equation*} R_{u u}=-\frac{\theta, u v}{\sin \theta}, \quad R_{u v}=-\frac{\cos \theta}{\sin \theta} \theta_{, u v} \tag{15.37} \end{equation*}(15.37)Ruu=θ,uvsinθ,Ruv=cosθsinθθ,uv
(b) Finally, show that
(15.38) R = 2 θ , u v sin θ (15.38) R = 2 θ , u v sin θ {:(15.38)R=-2(theta_(,uv))/(sin theta):}\begin{equation*} R=-2 \frac{\theta_{, u v}}{\sin \theta} \tag{15.38} \end{equation*}(15.38)R=2θ,uvsinθ
(c) If the spacetime is of constant curvature K K KKK, then it has the property that R μ ν = K g μ ν R μ ν = K g μ ν R_(mu nu)=Kg_(mu nu)R_{\mu \nu}=K g_{\mu \nu}Rμν=Kgμν. Show that this implies that we have
(15.39) θ , u v = α sin θ (15.39) θ , u v = α sin θ {:(15.39)theta_(,uv)=alpha sin theta:}\begin{equation*} \theta_{, u v}=\alpha \sin \theta \tag{15.39} \end{equation*}(15.39)θ,uv=αsinθ
where α α alpha\alphaα is a constant to be determined.
This is the equation of motion for the sineGordon field theory, which is widely used in quantum field theory.
(15.5) In quantum field theory, the vacuum has an energy density owing to the zero-point energy of the matter fields. This energy of often removed (or renormalized) from the theory upon quantization. Let's interpret the vacuum energy realistically, and see what the effect is on the cosmology of the Universe. The strong nuclear interaction operates at energies of order 1 GeV over characteristic lengths of 10 15 m 10 15 m 10^(-15)m10^{-15} \mathrm{~m}1015 m.
(a) Compute the characteristic energy density of the nuclear matter field.
(b) Make an order of magnitude estimate of the radius of curvature of a universe with this energy density and comment on the result.
(15.6) Consider a particle propagating radially in the de Sitter universe with Lagrangian
(15.40) L = [ ( d t d τ ) 2 a ( t ) 2 ( d r d τ ) 2 ] 1 2 (15.40) L = d t d τ 2 a ( t ) 2 d r d τ 2 1 2 {:(15.40)L=[((dt)/((d)tau))^(2)-a(t)^(2)(((d)r)/((d)tau))^(2)]^((1)/(2)):}\begin{equation*} L=\left[\left(\frac{\mathrm{d} t}{\mathrm{~d} \tau}\right)^{2}-a(t)^{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}\right]^{\frac{1}{2}} \tag{15.40} \end{equation*}(15.40)L=[(dt dτ)2a(t)2( dr dτ)2]12
where τ τ tau\tauτ is the proper time.
(a) By eliminating the dependence on d r / d τ d r / d τ dr//dtau\mathrm{d} r / \mathrm{d} \taudr/dτ in the equations of motion, show that the t t ttt coordinate obeys
(15.41) d 2 t d τ 2 + a ˙ a [ ( d t d τ ) 2 1 ] = 0 (15.41) d 2 t d τ 2 + a ˙ a d t d τ 2 1 = 0 {:(15.41)(d^(2)t)/((d)tau^(2))+((a^(˙)))/(a)[(((d)t)/((d)tau))^(2)-1]=0:}\begin{equation*} \frac{\mathrm{d}^{2} t}{\mathrm{~d} \tau^{2}}+\frac{\dot{a}}{a}\left[\left(\frac{\mathrm{~d} t}{\mathrm{~d} \tau}\right)^{2}-1\right]=0 \tag{15.41} \end{equation*}(15.41)d2t dτ2+a˙a[( dt dτ)21]=0
(b) Verify that this latter equation is solved by
(15.42) a ( t ) 2 [ ( d t d τ ) 2 1 ] = const. (15.42) a ( t ) 2 d t d τ 2 1 =  const.  {:(15.42)a(t)^(2)[(((d)t)/((d)tau))^(2)-1]=" const. ":}\begin{equation*} a(t)^{2}\left[\left(\frac{\mathrm{~d} t}{\mathrm{~d} \tau}\right)^{2}-1\right]=\text { const. } \tag{15.42} \end{equation*}(15.42)a(t)2[( dt dτ)21]= const. 
(c) Use the result from (b) and the relation g μ ν u μ u ν = 1 g μ ν u μ u ν = 1 g_(mu nu)u^(mu)u^(nu)=-1g_{\mu \nu} u^{\mu} u^{\nu}=-1gμνuμuν=1 to show that the 3-velocity of the
particle varies as | u | 2 a ( t ) 2 | u | 2 a ( t ) 2 | vec(u)|^(2)prop a(t)^(-2)|\vec{u}|^{2} \propto a(t)^{-2}|u|2a(t)2
This demonstrates that the magnitude of the momentum of a particle falls off due to the expansion of the Universe.
(15.7) Consider sending a light signal from the origin of the de Sitter universe at a time t s t s t_(s)t_{\mathrm{s}}ts, which is received at a time t r t r t_(r)t_{\mathrm{r}}tr at radial position \ell.
(a) Show that
(15.43) = 1 H ( e H t s e H t r ) (15.43) = 1 H e H t s e H t r {:(15.43)ℓ=(1)/(H)(e^(-Ht_(s))-e^(-Ht_(r))):}\begin{equation*} \ell=\frac{1}{H}\left(\mathrm{e}^{-H t_{\mathrm{s}}}-\mathrm{e}^{-H t_{\mathrm{r}}}\right) \tag{15.43} \end{equation*}(15.43)=1H(eHtseHtr)
(b) Explain why this implies the existence of position ℓ^(')\ell^{\prime} beyond which the light signal cannot prop-
agate owing to the expansion of the universe.
As a result of this, the de Sitter universe is said to have an event horizon. See Chapter 19 for the full story.
(15.8) Consider de Sitter spacetime, with metric line element
d s 2 = d t 2 + e 2 H t ( d x 2 + d y 2 + d z 2 ) d s 2 = d t 2 + e 2 H t d x 2 + d y 2 + d z 2 ds^(2)=-dt^(2)+e^(2Ht)((d)x^(2)+dy^(2)+dz^(2))\mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{e}^{2 H t}\left(\mathrm{~d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2}\right)ds2=dt2+e2Ht( dx2+dy2+dz2)
Show that the geodesics for massive particles are straight lines.

Robertson-Walker spaces

The universe ought to be presumed too vast to have any character
C. S. Peirce (1839-1914)
In the last chapter, we looked at two spatially flat universes. We now relax the geometrical constraint of spatial flatness. This will allow us to identify a large class of geometries that can serve as the basis for modelling universes. These will not yet be fully fledged models, since we are only treating the left-hand (geometrical) half of Einstein's equation. It will be the job of the next chapter to fill the universes with matter to complete the modelling process.
The geometries in this chapter have constant spatial curvature. 1 1 ^(1){ }^{1}1 We shall find that these geometries can be divided into three different types, according to whether the curvature is positive, zero, or negative. 2 2 ^(2){ }^{2}2 The three types of geometry share many features, perhaps the most dramatic is a consequence of time dependence: the three geometries can shrink to a point at some value of the time coordinate t t ttt. At this point the curvature of spacetime becomes infinite and we have a singularity in the geometry. Singularities have a number of worrying features for the description of Nature in terms of fields that will occupy us in several contexts in the remainder of this book. However, perhaps we should not be too jumpy: it is likely that our Universe emerged from an initial singularity, the Big Bang, and in identifying it in these geometries, we might well have taken a step closer to uncovering the origin of spacetime itself.

16.1 Spaces with constant curvature

In order to allow the spatial part of the metric to take different functional forms, we write the line element in the form, identified in the last chapter, demanded by homogeneity and isotropy:
(16.1) d s 2 = d t 2 + a ( t ) 2 γ i j d x i d x j (16.1) d s 2 = d t 2 + a ( t ) 2 γ i j d x i d x j {:(16.1)ds^(2)=-dt^(2)+a(t)^(2)gamma_(ij)dx^(i)dx^(j):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2} \gamma_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j} \tag{16.1} \end{equation*}(16.1)ds2=dt2+a(t)2γij dxi dxj
It's worth keeping in mind that the constraints on the geometry demanded by homogeneity and isotropy, rather than anything about general relativity, give us this metric. The flat spatial universes (i.e. Universes 0 and 1) turn out to be special cases of the more general spaces described by this metric.

16.1 Spaces with constant curvature 169 16.2 Three Robertson-Walker spaces 173
Exercises 179
1 1 ^(1){ }^{1}1 That is, the curvature is the same at all points in space and so can be characterized using a single number. This feature is examined further below and also in Exercises 16.5 and 15.3.
2 2 ^(2){ }^{2}2 Reminder: A simple example of a twodimensional surface with positive curvature is a ball. The circumference of a circle at fixed θ θ theta\thetaθ is 2 π r sin θ 2 π r sin θ 2pi r sin theta2 \pi r \sin \theta2πrsinθ, which is less than 2 π r 2 π r 2pi r2 \pi r2πr. As a result, flattening the surface onto a flat plane causes it to tear, with gaps left in its projection. In contrast, a surface with negative curvature has circles with circumference larger than 2 π r 2 π r 2pi r2 \pi r2πr, so that a projection onto a plane causes folds in the surface owing to the overlaps in the projection.
3 3 ^(3){ }^{3}3 This can only be built out of γ i j γ i j gamma_(ij)\gamma_{i j}γij and the only other 3 -tensor available that maintains homogeneity and isotropy, the Levi-Civita tensor, which has components ε i j k ε i j k epsi_(ijk)\varepsilon_{i j k}εijk. All other tensors at our disposal lead to a preferred direction and therefore violate isotropy. In fact, we find that we only need γ i j γ i j gamma_(ij)\gamma_{i j}γij.
4 4 ^(4){ }^{4}4 See Chapter 11.
\curvearrowright We discuss Gaussian curvature in detail in Chapter 30. For now, we can simply regard K K KKK as a constant whose form ultimately describes the possible classes of spacetime.
5 5 ^(5){ }^{5}5 This relies on the computation
γ i k ( γ i k γ j l γ i l γ j k ) = 3 γ j l γ j l = 2 γ j l γ i k γ i k γ j l γ i l γ j k = 3 γ j l γ j l = 2 γ j l gamma^(ik)(gamma_(ik)gamma_(jl)-gamma_(il)gamma_(jk))=3gamma_(jl)-gamma_(jl)=2gamma_(jl)\gamma^{i k}\left(\gamma_{i k} \gamma_{j l}-\gamma_{i l} \gamma_{j k}\right)=3 \gamma_{j l}-\gamma_{j l}=2 \gamma_{j l}γik(γikγjlγilγjk)=3γjlγjl=2γjl. (16.3)
6 6 ^(6){ }^{6}6 Be careful to note these are threedimensional expressions for the spatial part of this spacetime. They are not the spacetime expressions that feed into the Einstein equation.
7 7 ^(7){ }^{7}7 Note here that we're discussing space only, and not spacetime. In the last chapter (Exercise 15.3), we said that the de Sitter universe (Universe 1) was a spacetime with constant curvature There we described a spacetime of constant curvature as having a Riemann curvature tensor with components de termined by the Ricci scalar, and obeying the expression R μ ν = 1 4 R g μ ν R μ ν = 1 4 R g μ ν R_(mu nu)=(1)/(4)Rg_(mu nu)R_{\mu \nu}=\frac{1}{4} R g_{\mu \nu}Rμν=14Rgμν. An equivalent description is that a space time with curvature tensor has compo nents
R μ ν α β = C ( g μ α g ν β g μ β g ν α ) , R μ ν α β = C g μ α g ν β g μ β g ν α , R_(mu nu alpha beta)=C(g_(mu alpha)g_(nu beta)-g_(mu beta)g_(nu alpha)),R_{\mu \nu \alpha \beta}=C\left(g_{\mu \alpha} g_{\nu \beta}-g_{\mu \beta} g_{\nu \alpha}\right),Rμναβ=C(gμαgνβgμβgνα), with C C CCC a constant. (This is usually called K K KKK, but we distinguish it here from the value of the constant in the three-dimensional case to avoid any confusion.) The link between the two descriptions is made in the exercises for this chapter.
8 8 ^(8){ }^{8}8 See Exercise 16.4.
Consider the part of the geometry defined by the line element of threedimensional space d σ 2 = γ i j d x i d x j d σ 2 = γ i j d x i d x j dsigma^(2)=gamma_(ij)dx^(i)dx^(j)\mathrm{d} \sigma^{2}=\gamma_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j}dσ2=γij dxi dxj. Recall that this is the geometry of the three-dimensional, spacelike hypersurfaces we have constructed at each fixed value of the time coordinate t t ttt. We shall construct the threedimensional Riemann curvature tensor, denoted ( 3 ) R ( 3 ) R ^((3))R{ }^{(3)} \boldsymbol{R}(3)R, that describes the curvature of these slices of spacetime. 3 3 ^(3){ }^{3}3 The form of the components of ( 3 ) R ( 3 ) R ^((3))R{ }^{(3)} \boldsymbol{R}(3)R is fixed, not only by homogeneity and isotropy, but also by the usual symmetries of the Riemann tensor. 4 4 ^(4){ }^{4}4 There is only one arrangement that obeys all of these symmetries and it is written as
(16.2) ( 3 ) R i j k l = K ( γ i k γ j l γ i l γ j k ) , (16.2) ( 3 ) R i j k l = K γ i k γ j l γ i l γ j k , {:(16.2)^((3))R_(ijkl)=K(gamma_(ik)gamma_(jl)-gamma_(il)gamma_(jk))",":}\begin{equation*} { }^{(3)} R_{i j k l}=K\left(\gamma_{i k} \gamma_{j l}-\gamma_{i l} \gamma_{j k}\right), \tag{16.2} \end{equation*}(16.2)(3)Rijkl=K(γikγjlγilγjk),
where K K KKK is the Gaussian curvature: a constant with dimensions 1 / ( length ) 2 1 / (  length  ) 2 1//(" length ")^(2)1 /(\text { length })^{2}1/( length )2.
If we contract the components ( 3 ) R i j k l ( 3 ) R i j k l ^((3))R_(ijkl){ }^{(3)} R_{i j k l}(3)Rijkl using the 3 -metric γ i k γ i k gamma^(ik)\gamma^{i k}γik we obtain the components of the three-dimensional Ricci tensor, ( 3 ) R j l ( 3 ) R j l ^((3))R_(jl){ }^{(3)} R_{j l}(3)Rjl, which are 5 5 ^(5){ }^{5}5
(16.4) ( 3 ) R j l = 2 K γ j l . (16.4) ( 3 ) R j l = 2 K γ j l . {:(16.4)^((3))R_(jl)=2Kgamma_(jl).:}\begin{equation*} { }^{(3)} R_{j l}=2 K \gamma_{j l} . \tag{16.4} \end{equation*}(16.4)(3)Rjl=2Kγjl.
It follows that the three-dimensional Ricci scalar is ( 3 ) R = 6 K . 6 ( 3 ) R = 6 K . 6 ^((3))R=6K.^(6){ }^{(3)} R=6 K .{ }^{6}(3)R=6K.6
Having argued what the three-curvature must be from homogeneity, isotropy and the properties of the Riemann tensor, we next need to find the metric that gives this three-curvature. Robertson and Walker showed that there is only one metric that is consistent with the constraints. It is the one where the function γ i j γ i j gamma_(ij)\gamma_{i j}γij takes the form
(16.5) γ i j = [ 1 + K 4 ( x 2 + y 2 + z 2 ) ] 2 δ i j . (16.5) γ i j = 1 + K 4 x 2 + y 2 + z 2 2 δ i j . {:(16.5)gamma_(ij)=[1+(K)/(4)(x^(2)+y^(2)+z^(2))]^(-2)delta_(ij).:}\begin{equation*} \gamma_{i j}=\left[1+\frac{K}{4}\left(x^{2}+y^{2}+z^{2}\right)\right]^{-2} \delta_{i j} . \tag{16.5} \end{equation*}(16.5)γij=[1+K4(x2+y2+z2)]2δij.
Spaces with this metric are called spaces of constant curvature. 7 7 ^(7){ }^{7}7
Example 16.1
The metric line element using the above form of γ i j γ i j gamma_(ij)\gamma_{i j}γij is written as
(16.7) d s 2 = d t 2 + a ( t ) 2 ( d x 2 + d y 2 + d z 2 ) [ 1 + K 4 ( x 2 + y 2 + z 2 ) ] 2 (16.7) d s 2 = d t 2 + a ( t ) 2 d x 2 + d y 2 + d z 2 1 + K 4 x 2 + y 2 + z 2 2 {:(16.7)ds^(2)=-dt^(2)+a(t)^(2)(((d)x^(2)+dy^(2)+dz^(2)))/([1+(K)/(4)(x^(2)+y^(2)+z^(2))]^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2} \frac{\left(\mathrm{~d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2}\right)}{\left[1+\frac{K}{4}\left(x^{2}+y^{2}+z^{2}\right)\right]^{2}} \tag{16.7} \end{equation*}(16.7)ds2=dt2+a(t)2( dx2+dy2+dz2)[1+K4(x2+y2+z2)]2
Written in polar coordinates, this becomes
(16.8) d s 2 = d t 2 + a ( t ) 2 ( d R 2 + R 2 d Ω 2 ) ( 1 + K R 2 4 ) 2 , (16.8) d s 2 = d t 2 + a ( t ) 2 d R 2 + R 2 d Ω 2 1 + K R 2 4 2 , {:(16.8)ds^(2)=-dt^(2)+a(t)^(2)(((d)R^(2)+R^(2)(d)Omega^(2)))/((1+(KR^(2))/(4))^(2))",":}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2} \frac{\left(\mathrm{~d} R^{2}+R^{2} \mathrm{~d} \Omega^{2}\right)}{\left(1+\frac{K R^{2}}{4}\right)^{2}}, \tag{16.8} \end{equation*}(16.8)ds2=dt2+a(t)2( dR2+R2 dΩ2)(1+KR24)2,
where d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}dΩ2=dθ2+sin2θ dϕ2 and the radial coordinate here is written as R ( = R ( = R(=R(=R(= x 2 + y 2 + z 2 ) x 2 + y 2 + z 2 {:sqrt(x^(2)+y^(2)+z^(2)))\left.\sqrt{x^{2}+y^{2}+z^{2}}\right)x2+y2+z2). This expression can be simplified if we exploit the freedom to choose a new radial coordinate r r rrr. This choice is constrained by our wanting the proper circumference that looks like a ( t ) × ( 2 π r ) a ( t ) × ( 2 π r ) a(t)xx(2pi r)a(t) \times(2 \pi r)a(t)×(2πr), which we can achieve if the θ θ theta\thetaθ dependent term in the metric has the form a 2 r 2 d θ 2 a 2 r 2 d θ 2 a^(2)r^(2)dtheta^(2)a^{2} r^{2} \mathrm{~d} \theta^{2}a2r2 dθ2. We are therefore motivated to choose
(16.9) r = R 1 + K R 2 4 (16.9) r = R 1 + K R 2 4 {:(16.9)r=(R)/(1+(KR^(2))/(4)):}\begin{equation*} r=\frac{R}{1+\frac{K R^{2}}{4}} \tag{16.9} \end{equation*}(16.9)r=R1+KR24
Substituting for this coordinate, we find, after some algebra, 8 8 ^(8){ }^{8}8 that
(16.10) d s 2 = d t 2 + a ( t ) 2 ( d r 2 1 K r 2 + r 2 d Ω 2 ) (16.10) d s 2 = d t 2 + a ( t ) 2 d r 2 1 K r 2 + r 2 d Ω 2 {:(16.10)ds^(2)=-dt^(2)+a(t)^(2)(((d)r^(2))/(1-Kr^(2))+r^(2)(d)Omega^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left(\frac{\mathrm{~d} r^{2}}{1-K r^{2}}+r^{2} \mathrm{~d} \Omega^{2}\right) \tag{16.10} \end{equation*}(16.10)ds2=dt2+a(t)2( dr21Kr2+r2 dΩ2)
The resulting Robertson-Walker metric is written in terms of a line element as
(16.11) d s 2 = d t 2 + a ( t ) 2 [ d r 2 1 K r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (16.11) d s 2 = d t 2 + a ( t ) 2 d r 2 1 K r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.11)ds^(2)=-dt^(2)+a(t)^(2)[((d)r^(2))/(1-Kr^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-K r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{16.11} \end{equation*}(16.11)ds2=dt2+a(t)2[ dr21Kr2+r2( dθ2+sin2θ dϕ2)]
Note that this metric has the useful property that if the Gaussian curvature vanishes, i.e. if K = 0 K = 0 K=0K=0K=0, it reduces back to the flat spacetime of the de Sitter model (or Universe 1) of the previous chapter.
The Gaussian curvature can be zero, but can also take positive or negative values. If it is positive, the geometry is spherical (the Gaussian curvature of a sphere of radius r r rrr is K = + 1 / r 2 K = + 1 / r 2 K=+1//r^(2)K=+1 / r^{2}K=+1/r2 ); if it is negative, the geometry is hyperbolic. We will deal with these geometrical aspects in more detail below (and also in Part V), but for
now we would like to simplify the metric a bit further by rescaling variables so that
rescaled curvature k k kkk can only take three values: k = 0 k = 0 k=0k=0k=0 (flat), k = + 1 k = + 1 k=+1k=+1k=+1 (spherical) and rescaled curvature k k kkk can only take three values: k = 0 k = 0 k=0k=0k=0 (flat), k = + 1 k = + 1 k=+1k=+1k=+1 (spherical) and
k = 1 k = 1 k=-1k=-1k=1 (hyperbolic). We do this by setting k = K / | K | k = K / | K | k=K//|K|k=K /|K|k=K/|K| if K 0 K 0 K!=0K \neq 0K0, and setting k = 0 k = 0 k=0k=0k=0 if K = 0 K = 0 K=0K=0K=0. Next we introduce a rescaled, dimensionless radial coordinate r ¯ = | K | 1 2 r r ¯ = | K | 1 2 r bar(r)=|K|^((1)/(2))r\bar{r}=|K|^{\frac{1}{2}} rr¯=|K|12r, so we have
(16.12) d s 2 = d t 2 + a ( t ) 2 | K | [ d r ¯ 2 1 k r ¯ 2 + r ¯ 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (16.12) d s 2 = d t 2 + a ( t ) 2 | K | d r ¯ 2 1 k r ¯ 2 + r ¯ 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.12)ds^(2)=-dt^(2)+(a(t)^(2))/(|K|)[(d bar(r)^(2))/(1-k bar(r)^(2))+ bar(r)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\frac{a(t)^{2}}{|K|}\left[\frac{\mathrm{d} \bar{r}^{2}}{1-k \bar{r}^{2}}+\bar{r}^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{16.12} \end{equation*}(16.12)ds2=dt2+a(t)2|K|[dr¯21kr¯2+r¯2( dθ2+sin2θ dϕ2)]
The part in the square brackets is now dimensionless and so the prefactor has dimensions (length) 2 2 ^(2){ }^{2}2. Finally, we redefine the scale factor a ( t ) a ( t ) a(t)a(t)a(t) so that it becomes
(16.13) a ( t ) := { a ( t ) / | K | K 0 a ( t ) K = 0 (16.13) a ( t ) := a ( t ) / | K | K 0 a ( t ) K = 0 {:(16.13)a(t):={[a(t)//sqrt(|K|),K!=0],[a(t),K=0]:}:}a(t):=\left\{\begin{array}{cc} a(t) / \sqrt{|K|} & K \neq 0 \tag{16.13}\\ a(t) & K=0 \end{array}\right.(16.13)a(t):={a(t)/|K|K0a(t)K=0
This implies that a ( t ) a ( t ) a(t)a(t)a(t), which was previously a dimensionless scaling factor, has taken on the dimensions of length. The curvature of the universe is now given by K = K = K=K=K= k / a ( t ) 2 k / a ( t ) 2 k//a(t)^(2)k / a(t)^{2}k/a(t)2. Motivated by the result that Gaussian curvature of a sphere is K = 1 / r 2 K = 1 / r 2 K=1//r^(2)K=1 / r^{2}K=1/r2, the factor a ( t ) a ( t ) a(t)a(t)a(t) is sometimes known as the radius of the universe in this context.
Dropping the bar from r ¯ r ¯ bar(r)\bar{r}r¯ in the last example (purely to save on writing), we have the final result for the Robertson-Walker metric
(16.14) d s 2 = d t 2 + a ( t ) 2 [ d r 2 1 k r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (16.14) d s 2 = d t 2 + a ( t ) 2 d r 2 1 k r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.14)ds^(2)=-dt^(2)+a(t)^(2)[((d)r^(2))/(1-kr^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-k r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{16.14} \end{equation*}(16.14)ds2=dt2+a(t)2[ dr21kr2+r2( dθ2+sin2θ dϕ2)]
where k = 1 , 0 k = 1 , 0 k=1,0k=1,0k=1,0 or -1 .
In order to visualize the three possibilities presented by the values of the curvature parameter k k kkk (which may be ± 1 ± 1 +-1\pm 1±1 or 0 ), we shall, in Section 16.2, consider the result of embedding the 3 -spaces described by line element d σ 2 d σ 2 dsigma^(2)\mathrm{d} \sigma^{2}dσ2 into four-dimensional flat space.

Example 16.2

Before we get there, we can make some progress by noting the similarity between the Robertson-Walker spaces and the 2-spaces discussed in Appendix D. There we present an example analogous to the k = 1 k = 1 k=1k=1k=1 Robertson-Walker space, which is a sphere. We also meet an example analogous to the k = 1 k = 1 k=-1k=-1k=1 Robertson-Walker space case, which can't, as a whole, be embedded in Euclidean space, but can be represented as a 2 -sheet hyperboloid when embedded in Minkowski space. As might be expected, we shall see below how we can also characterize the k = 1 k = 1 k=1k=1k=1 RobertsonWalker case as a sort of spherical space, and the k = 1 k = 1 k=-1k=-1k=1 case as a hyperbolic space. The k = 0 k = 0 k=0k=0k=0 case is a flat space. Also similar to what we present in Appendix D, the factor ( 1 k r 2 ) 1 k r 2 (1-kr^(2))\left(1-k r^{2}\right)(1kr2) can be removed from eqn 16.14 if we replace the (now dimensionless) variable r r rrr with a convenient alternative variable χ , 9 χ , 9 chi,^(9)\chi,{ }^{9}χ,9 which leads to the substitution d χ 2 = d r 2 / ( 1 k r 2 ) d χ 2 = d r 2 / 1 k r 2 dchi^(2)=dr^(2)//(1-kr^(2))\mathrm{d} \chi^{2}=\mathrm{d} r^{2} /\left(1-k r^{2}\right)dχ2=dr2/(1kr2).
Robertson-Walker coordinates are comoving, so a galaxy has fixed ( r , θ , ϕ ) ( r , θ , ϕ ) (r,theta,phi)(r, \theta, \phi)(r,θ,ϕ). We have metric components
g t t = 1 , g r r = a 2 1 k r 2 g θ θ = a 2 r 2 , g ϕ ϕ = a 2 r 2 sin 2 θ . g t t = 1 , g r r = a 2 1 k r 2 g θ θ = a 2 r 2 , g ϕ ϕ = a 2 r 2 sin 2 θ . {:[g_(tt)=-1","quadg_(rr)=(a^(2))/(1-kr^(2))],[g_(theta theta)=a^(2)r^(2)","quadg_(phi phi)=a^(2)r^(2)sin^(2)theta.]:}\begin{gathered} g_{t t}=-1, \quad g_{r r}=\frac{a^{2}}{1-k r^{2}} \\ g_{\theta \theta}=a^{2} r^{2}, \quad g_{\phi \phi}=a^{2} r^{2} \sin ^{2} \theta . \end{gathered}gtt=1,grr=a21kr2gθθ=a2r2,gϕϕ=a2r2sin2θ.
Vielbein components are given by
( e t ) t ^ = 1 , ( e r ) r ^ = a ( 1 k r 2 ) 1 2 e t t ^ = 1 , e r r ^ = a 1 k r 2 1 2 (e_(t))^( hat(t))=1,quad(e_(r))^( hat(r))=(a)/((1-kr^(2))^((1)/(2)))\left(\boldsymbol{e}_{t}\right)^{\hat{t}}=1, \quad\left(\boldsymbol{e}_{r}\right)^{\hat{r}}=\frac{a}{\left(1-k r^{2}\right)^{\frac{1}{2}}}(et)t^=1,(er)r^=a(1kr2)12,
( e θ ) θ ^ = a r , ( e ϕ ) ϕ ^ = a r sin θ e θ θ ^ = a r , e ϕ ϕ ^ = a r sin θ (e_(theta))^( hat(theta))=ar,quad(e_(phi))^( hat(phi))=ar sin theta\left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}}=a r, \quad\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}=a r \sin \theta(eθ)θ^=ar,(eϕ)ϕ^=arsinθ.
The connection coefficients are
Γ r r t = a a ˙ 1 k r 2 , Γ t θ θ = a a ˙ r 2 , Γ t ϕ ϕ = a a ˙ r 2 sin 2 θ , Γ r r r = k r 1 k r 2 , Γ t j i = a ˙ a δ i j , Γ r Γ θ θ r = r ( 1 k r 2 ) , Γ ϕ ϕ θ = r ( 1 k r 2 ) sin 2 θ , Γ r θ θ = Γ ϕ r ϕ = 1 r , Γ ϕ ϕ ϕ = sin θ cos θ , = cot θ . Γ r r t = a a ˙ 1 k r 2 , Γ t θ θ = a a ˙ r 2 , Γ t ϕ ϕ = a a ˙ r 2 sin 2 θ , Γ r r r = k r 1 k r 2 , Γ t j i = a ˙ a δ i j , Γ r Γ θ θ r = r 1 k r 2 , Γ ϕ ϕ θ = r 1 k r 2 sin 2 θ , Γ r θ θ = Γ ϕ r ϕ = 1 r , Γ ϕ ϕ ϕ = sin θ cos θ , = cot θ . {:[Gamma_(rr)^(t)=(a(a^(˙)))/(1-kr^(2))","],[Gamma^(t)_(theta theta)=aa^(˙)r^(2)","],[Gamma^(t)_(phi phi)=aa^(˙)r^(2)sin^(2)theta","],[Gamma_(rr)^(r)=(kr)/(1-kr^(2))","],[Gamma_(tj)^(i)=((a^(˙)))/(a)delta^(i)_(j)","],[Gamma^(r)],[Gamma_(theta theta)^(r)=-r(1-kr^(2))","],[Gamma_(phi phi)^(theta)=-r(1-kr^(2))sin^(2)theta","],[Gamma_(r theta)^(theta)=Gamma^(phi)_(r phi)=(1)/(r)","],[Gamma_(phi phi)^(phi)=-sin theta cos theta","],[=cot theta.]:}\begin{aligned} \Gamma_{r r}^{t} & =\frac{a \dot{a}}{1-k r^{2}}, \\ \Gamma^{t}{ }_{\theta \theta} & =a \dot{a} r^{2}, \\ \Gamma^{t}{ }_{\phi \phi} & =a \dot{a} r^{2} \sin ^{2} \theta, \\ \Gamma_{r r}^{r} & =\frac{k r}{1-k r^{2}}, \\ \Gamma_{t j}^{i} & =\frac{\dot{a}}{a} \delta^{i}{ }_{j}, \\ \Gamma^{r} & \\ \Gamma_{\theta \theta}^{r} & =-r\left(1-k r^{2}\right), \\ \Gamma_{\phi \phi}^{\theta} & =-r\left(1-k r^{2}\right) \sin ^{2} \theta, \\ \Gamma_{r \theta}^{\theta} & =\Gamma^{\phi}{ }_{r \phi}=\frac{1}{r}, \\ \Gamma_{\phi \phi}^{\phi} & =-\sin \theta \cos \theta, \\ & =\cot \theta . \end{aligned}Γrrt=aa˙1kr2,Γtθθ=aa˙r2,Γtϕϕ=aa˙r2sin2θ,Γrrr=kr1kr2,Γtji=a˙aδij,ΓrΓθθr=r(1kr2),Γϕϕθ=r(1kr2)sin2θ,Γrθθ=Γϕrϕ=1r,Γϕϕϕ=sinθcosθ,=cotθ.
The Ricci tensor has non-zero components in the orthonormal frame of
R i t ^ = 3 a ¨ a , R i t ^ ^ = a ¨ a + 2 ( a ˙ a ) 2 + 2 k a , R i t ^ = 3 a ¨ a , R i t ^ ^ = a ¨ a + 2 a ˙ a 2 + 2 k a , {:[R_(i hat(t))=-3((a^(¨)))/(a)","],[R_( hat(it) hat())=((a^(¨)))/(a)+2(((a^(˙)))/(a))^(2)+2(k)/(a)","]:}\begin{aligned} & R_{i \hat{t}}=-3 \frac{\ddot{a}}{a}, \\ & R_{\hat{i t} \hat{}}=\frac{\ddot{a}}{a}+2\left(\frac{\dot{a}}{a}\right)^{2}+2 \frac{k}{a}, \end{aligned}Rit^=3a¨a,Rit^^=a¨a+2(a˙a)2+2ka,
The Ricci scalar is
R = 6 ( a ¨ a + a ˙ 2 a 2 + k a 2 ) . R = 6 a ¨ a + a ˙ 2 a 2 + k a 2 . R=6(((a^(¨)))/(a)+(a^(˙)^(2))/(a^(2))+(k)/(a^(2))).R=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}+\frac{k}{a^{2}}\right) .R=6(a¨a+a˙2a2+ka2).
9 9 ^(9){ }^{9}9 In Appendix D, it is shown that for the k = 1 k = 1 k=1k=1k=1 case, r = sin χ r = sin χ r=sin chir=\sin \chir=sinχ simplifies the line element. The corresponding choice for k = 1 k = 1 k=-1k=-1k=1 is r = sinh χ r = sinh χ r=sinh chir=\sinh \chir=sinhχ, while we simply take r = χ r = χ r=chir=\chir=χ for k = 0 k = 0 k=0k=0k=0.
The metric has non-zero components
g t t = 1 , g χ χ = a ( t ) 2 , g θ θ = a ( t ) 2 Σ ( χ ) 2 , g ϕ ϕ = a ( t ) 2 Σ ( χ ) 2 sin 2 θ . g t t = 1 , g χ χ = a ( t ) 2 , g θ θ = a ( t ) 2 Σ ( χ ) 2 , g ϕ ϕ = a ( t ) 2 Σ ( χ ) 2 sin 2 θ . {:[g_(tt)=-1","],[g_(chi chi)=a(t)^(2)","],[g_(theta theta)=a(t)^(2)Sigma(chi)^(2)","],[g_(phi phi)=a(t)^(2)Sigma(chi)^(2)sin^(2)theta.]:}\begin{aligned} g_{t t} & =-1, \\ g_{\chi \chi} & =a(t)^{2}, \\ g_{\theta \theta} & =a(t)^{2} \Sigma(\chi)^{2}, \\ g_{\phi \phi} & =a(t)^{2} \Sigma(\chi)^{2} \sin ^{2} \theta . \end{aligned}gtt=1,gχχ=a(t)2,gθθ=a(t)2Σ(χ)2,gϕϕ=a(t)2Σ(χ)2sin2θ.
We have vielbein components
( e t ) t ^ = 1 e t t ^ = 1 (e_(t))^( hat(t))=1\left(e_{t}\right)^{\hat{t}}=1(et)t^=1,
( e χ ) χ ^ = a ( t ) e χ χ ^ = a ( t ) (e_(chi))^( hat(chi))=a(t)\left(e_{\chi}\right)^{\hat{\chi}}=a(t)(eχ)χ^=a(t),
( e θ ) θ ^ = a ( t ) Σ ( χ ) e θ θ ^ = a ( t ) Σ ( χ ) (e_(theta))^( hat(theta))=a(t)Sigma(chi)\left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}}=a(t) \Sigma(\chi)(eθ)θ^=a(t)Σ(χ),
( e ϕ ) ϕ ^ = a ( t ) Σ ( χ ) sin θ e ϕ ϕ ^ = a ( t ) Σ ( χ ) sin θ (e_(phi))^( hat(phi))=a(t)Sigma(chi)sin theta\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}=a(t) \Sigma(\chi) \sin \theta(eϕ)ϕ^=a(t)Σ(χ)sinθ.
10 10 ^(10){ }^{10}10 Recall that comoving coordinates must have g 00 , i = g 0 i , 0 = 0 g 00 , i = g 0 i , 0 = 0 g_(00,i)=g_(0i,0)=0g_{00, i}=g_{0 i, 0}=0g00,i=g0i,0=0, which these certainly do.
In terms of the new dimensionless variable χ χ chi\chiχ from the last example, we are able to summarize the Robertson-Walker spaces in a form that's very useful for applications, where the coordinates resemble ordinary spherical coordinates. These are given in the box below.
In spherical coordinates, the three Robertson-Walker spacetimes are written as
(16.15) d s 2 = d t 2 + a ( t ) 2 [ d χ 2 + Σ ( χ ) 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (16.15) d s 2 = d t 2 + a ( t ) 2 d χ 2 + Σ ( χ ) 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.15)ds^(2)=-dt^(2)+a(t)^(2)[(d)chi^(2)+Sigma(chi)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\mathrm{~d} \chi^{2}+\Sigma(\chi)^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{16.15} \end{equation*}(16.15)ds2=dt2+a(t)2[ dχ2+Σ(χ)2( dθ2+sin2θ dϕ2)]
where
(16.16) Σ ( χ ) = sin χ ( k = + 1 ) Σ ( χ ) = χ ( k = 0 ) Σ ( χ ) = sinh χ ( k = 1 ) (16.16) Σ ( χ ) = sin χ ( k = + 1 ) Σ ( χ ) = χ ( k = 0 ) Σ ( χ ) = sinh χ ( k = 1 ) {:(16.16){:[Sigma(chi)=sin chi,(k=+1)],[Sigma(chi)=chi,(k=0)],[Sigma(chi)=sinh chi,(k=-1)]:}:}\begin{array}{cc} \Sigma(\chi)=\sin \chi & (k=+1) \\ \Sigma(\chi)=\chi & (k=0) \tag{16.16}\\ \Sigma(\chi)=\sinh \chi & (k=-1) \end{array}(16.16)Σ(χ)=sinχ(k=+1)Σ(χ)=χ(k=0)Σ(χ)=sinhχ(k=1)
The spatial hypersurfaces are swept out with coordinates
(16.17) 0 χ π , 0 θ π , π ϕ π , 0 χ < , 0 θ π , π ϕ π , 0 0 , 1 ) (16.17) 0 χ π , 0 θ π , π ϕ π , 0 χ < , 0 θ π , π ϕ π , 0 0 , 1 ) {:(16.17){:[0 <= chi <= pi",",0 <= theta <= pi",",-pi <= phi <= pi","],[0 <= chi < oo",",0 <= theta <= pi",",-pi <= phi <= pi","],[0 <= 0","-1)]:}:}\begin{array}{cll} 0 \leq \chi \leq \pi, & 0 \leq \theta \leq \pi, & -\pi \leq \phi \leq \pi, \tag{16.17}\\ 0 \leq \chi<\infty, & 0 \leq \theta \leq \pi, & -\pi \leq \phi \leq \pi, \\ 0 \leq 0,-1) \end{array}(16.17)0χπ,0θπ,πϕπ,0χ<,0θπ,πϕπ,00,1)
Notice the difference in the range of χ χ chi\chiχ for k = 1 k = 1 k=1k=1k=1 compared to the other cases. This tells us that the properties of k = 1 k = 1 k=1k=1k=1 space are rather different to the others. Of course, to understand the effect of a metric field we insert it into the Einstein equation. For this purpose we extract the components of the Riemann tensor R R R\boldsymbol{R}R and use these to compute the Ricci tensor and scalar. These are given in the margin on the previous page.
The Robertson-Walker metric is written in terms of comoving coordinates: the time t t ttt is the cosmic time variable, and the spatial variables are carried along with galaxies. 10 10 ^(10){ }^{10}10 In the next example, we deal with the proper length as a measure of the spacelike interval between two points on a spacelike hypersurface. This interval varies with time (so is different on different hypersurfaces) owing to the presence of the radius factor a ( t ) a ( t ) a(t)a(t)a(t). It also differs for different choices of the constant k k kkk.
Example 16.3
Consider two points along a radial line: ( r , θ , ϕ ) ( r , θ , ϕ ) (r,theta,phi)(r, \theta, \phi)(r,θ,ϕ) and ( r + d r , θ , ϕ ) ( r + d r , θ , ϕ ) (r+dr,theta,phi)(r+\mathrm{d} r, \theta, \phi)(r+dr,θ,ϕ). The element of proper distance along the line between the points is
(16.18) d l = a ( t ) d r ( 1 k r 2 ) 1 2 (16.18) d l = a ( t ) d r 1 k r 2 1 2 {:(16.18)dl=a(t)(dr)/((1-kr^(2))^((1)/(2))):}\begin{equation*} \mathrm{d} l=a(t) \frac{\mathrm{d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}} \tag{16.18} \end{equation*}(16.18)dl=a(t)dr(1kr2)12
If the two points are separated by fixed radial comoving coordinate ξ ξ xi\xiξ, they have a total spacelike interval between them of
(16.19) l ( t ) = 0 ξ a ( t ) d r ( 1 k r 2 ) 1 2 , (16.19) l ( t ) = 0 ξ a ( t ) d r 1 k r 2 1 2 , {:(16.19)l(t)=int_(0)^(xi)a(t)(dr)/((1-kr^(2))^((1)/(2)))",":}\begin{equation*} l(t)=\int_{0}^{\xi} a(t) \frac{\mathrm{d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}}, \tag{16.19} \end{equation*}(16.19)l(t)=0ξa(t)dr(1kr2)12,
which, on integration with the different choices of k k kkk, gives intervals
(16.20) l ( t ) = { a ( t ) sin 1 ξ ( k = 1 ) , a ( t ) ξ ( k = 0 ) , a ( t ) sinh 1 ξ ( k = 1 ) . (16.20) l ( t ) = a ( t ) sin 1 ξ ( k = 1 ) , a ( t ) ξ ( k = 0 ) , a ( t ) sinh 1 ξ ( k = 1 ) . {:(16.20)l(t)={[a(t)sin^(-1)xi,(k=1)","],[a(t)xi,(k=0)","],[a(t)sinh^(-1)xi,(k=-1).]:}:}l(t)=\left\{\begin{array}{cc} a(t) \sin ^{-1} \xi & (k=1), \tag{16.20}\\ a(t) \xi & (k=0), \\ a(t) \sinh ^{-1} \xi & (k=-1) . \end{array}\right.(16.20)l(t)={a(t)sin1ξ(k=1),a(t)ξ(k=0),a(t)sinh1ξ(k=1).
These proper distances are shown in the Fig. 16.1. Notice that for k = 1 k = 1 k=1k=1k=1 the proper distance reaches a finite limit for ξ 1 ξ 1 xi rarr1\xi \rightarrow 1ξ1 where sin 1 ξ = π / 2 sin 1 ξ = π / 2 sin^(-1)xi=pi//2\sin ^{-1} \xi=\pi / 2sin1ξ=π/2.
We can also define a proper radial velocity, measuring the rate of change of the interval, via the time derivative of l ( t ) l ( t ) l(t)l(t)l(t), from which we have
(16.21) d l d t = { d a d t sin 1 ξ ( k = 1 ) d a d t ξ ( k = 0 ) d a d t sinh 1 ξ ( k = 1 ) (16.21) d l d t = d a d t sin 1 ξ ( k = 1 ) d a d t ξ ( k = 0 ) d a d t sinh 1 ξ ( k = 1 ) {:(16.21)(dl)/((d)t)={[(da)/((d)t)sin^(-1)xi,(k=1)],[(da)/((d)t)xi,(k=0)],[(da)/((d)t)*sinh^(-1)xi,(k=-1)]:}:}\frac{\mathrm{d} l}{\mathrm{~d} t}=\left\{\begin{array}{cc} \frac{\mathrm{d} a}{\mathrm{~d} t} \sin ^{-1} \xi & (k=1) \tag{16.21}\\ \frac{\mathrm{d} a}{\mathrm{~d} t} \xi & (k=0) \\ \frac{\mathrm{d} a}{\mathrm{~d} t} \cdot \sinh ^{-1} \xi & (k=-1) \end{array}\right.(16.21)dl dt={da dtsin1ξ(k=1)da dtξ(k=0)da dtsinh1ξ(k=1)
Notice how the part that depends on ξ ξ xi\xiξ can be replaced by l ( t ) / a ( t ) l ( t ) / a ( t ) l(t)//a(t)l(t) / a(t)l(t)/a(t) in each case. Following from this, we have the result for the proper length that
(16.22) d l ( t ) d t = 1 a ( t ) d a ( t ) d t l ( t ) (16.22) d l ( t ) d t = 1 a ( t ) d a ( t ) d t l ( t ) {:(16.22)(dl(t))/(dt)=(1)/(a(t))(da(t))/(dt)l(t):}\begin{equation*} \frac{\mathrm{d} l(t)}{\mathrm{d} t}=\frac{1}{a(t)} \frac{\mathrm{d} a(t)}{\mathrm{d} t} l(t) \tag{16.22} \end{equation*}(16.22)dl(t)dt=1a(t)da(t)dtl(t)
The equation describes the change in the proper spacelike interval as a function of time and therefore describes the expansion of space itself. The equation is usually rewritten in terms of a proper velocity v p = d l ( t ) / d t v p = d l ( t ) / d t v_(p)=dl(t)//dtv_{\mathrm{p}}=\mathrm{d} l(t) / \mathrm{d} tvp=dl(t)/dt, and proper distance d p = l ( t ) d p = l ( t ) d_(p)=l(t)d_{\mathrm{p}}=l(t)dp=l(t) as Hubble's law
(16.23) v p = H ( t ) d p (16.23) v p = H ( t ) d p {:(16.23)v_(p)=H(t)d_(p):}\begin{equation*} v_{\mathrm{p}}=H(t) d_{\mathrm{p}} \tag{16.23} \end{equation*}(16.23)vp=H(t)dp
where the function H ( t ) H ( t ) H(t)H(t)H(t) is the Hubble parameter H ( t ) = 1 a d a d t H ( t ) = 1 a d a d t H(t)=(1)/(a)((d)a)/((d)t)H(t)=\frac{1}{a} \frac{\mathrm{~d} a}{\mathrm{~d} t}H(t)=1a da dt, as introduced in Chapter 15.
These notions of proper distance are useful in giving a more complete geometrical description of the Robertson-Walker spaces, which is our next topic.

16.2 Three Robertson-Walker spaces

The Robertson-Walker metric provides us with three classes of expanding spacetime with line element
(16.24) d s 2 = d t 2 + a ( t ) 2 [ d r 2 1 k r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (16.24) d s 2 = d t 2 + a ( t ) 2 d r 2 1 k r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.24)ds^(2)=-dt^(2)+a(t)^(2)[((d)r^(2))/(1-kr^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-k r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{16.24} \end{equation*}(16.24)ds2=dt2+a(t)2[ dr21kr2+r2( dθ2+sin2θ dϕ2)]
The three spaces are characterized by the choice of curvature parameters k = 0 k = 0 k=0k=0k=0 or k = ± 1 k = ± 1 k=+-1k= \pm 1k=±1. Next, we examine each of the spaces in turn in terms of their geometrical properties. To do this we compare the proper circumference of a circle to its proper radius. This yields 2 π 2 π 2pi2 \pi2π if the space is flat, but differs from this value if it is not. The idea is exactly that of Sidenote 2, that on a surface with positive curvature the ratio of circumference to radius is less than 2 π 2 π 2pi2 \pi2π (so the surface tears when projected onto a flat plane), whereas for a negative curvature surface the ratio is more than 2 π 2 π 2pi2 \pi2π (and the surface folds up as it's projected onto a flat plane). Recall that a galaxy at radial coordinate ξ ξ xi\xiξ is separated from a galaxy at the origin by a spacelike interval l ( t ) = a ( t ) ξ l ( t ) = a ( t ) ξ l(t)=a(t)xil(t)=a(t) \xil(t)=a(t)ξ. Taking θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2, the integral of the line element over a circle of coordinate radius ξ ξ xi\xiξ gives a circumference 2 π a ( t ) ξ 2 π a ( t ) ξ 2pi a(t)xi2 \pi a(t) \xi2πa(t)ξ for each of the spaces.
The case k = 0 k = 0 k=0k=0k=0
We describe k = 0 k = 0 k=0k=0k=0 universes as flat. The line element is given by
(16.25) d s 2 = d t 2 + a ( t ) 2 [ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (16.25) d s 2 = d t 2 + a ( t ) 2 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.25)ds^(2)=-dt^(2)+a(t)^(2)[(d)r^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{16.25} \end{equation*}(16.25)ds2=dt2+a(t)2[ dr2+r2( dθ2+sin2θ dϕ2)]
This is identical to Universe 1 discussed in the last chapter. If k = 0 k = 0 k=0k=0k=0 and a ( t ) a ( t ) a(t)a(t)a(t) is a constant, then we have a Riemann tensor R = 0 R = 0 R=0\boldsymbol{R}=0R=0 for all time
Fig. 16.1 The ratio of the proper distance l ( t ) l ( t ) l(t)l(t)l(t) to the expansion factor a ( t ) a ( t ) a(t)a(t)a(t) as a function of radial coordinate ξ ξ xi\xiξ.
11 11 ^(11){ }^{11}11 The embedding required is
w = a cos χ , x = a sin χ sin θ cos ϕ , y = a sin χ sin θ sin ϕ , z = a sin χ cos θ . w = a cos χ , x = a sin χ sin θ cos ϕ , y = a sin χ sin θ sin ϕ , z = a sin χ cos θ . {:[w=a cos chi","],[x=a sin chi sin theta cos phi","],[y=a sin chi sin theta sin phi","],[z=a sin chi cos theta.]:}\begin{aligned} w & =a \cos \chi, \\ x & =a \sin \chi \sin \theta \cos \phi, \\ y & =a \sin \chi \sin \theta \sin \phi, \\ z & =a \sin \chi \cos \theta . \end{aligned}w=acosχ,x=asinχsinθcosϕ,y=asinχsinθsinϕ,z=asinχcosθ.

(16.28)

Fig. 16.2 (a) The three-dimensional k = 1 k = 1 k=1k=1k=1 (closed) universe in fourdimensional Euclidean space, with one rotational degree of freedom suppressed. The embedding is w 2 + x 2 + w 2 + x 2 + w^(2)+x^(2)+w^{2}+x^{2}+w2+x2+ y 2 + z 2 = a 2 y 2 + z 2 = a 2 y^(2)+z^(2)=a^(2)y^{2}+z^{2}=a^{2}y2+z2=a2 (the equation of a fourdimensional sphere). (b) The threedimensional k = 1 k = 1 k=-1k=-1k=1 (hyperbolic) universe in four-dimensional Minkowski space with one rotational degree of freedom suppressed. The embedding is w 2 x 2 y 2 z 2 = a 2 w 2 x 2 y 2 z 2 = a 2 w^(2)-x^(2)-y^(2)-z^(2)=a^(2)w^{2}-x^{2}-y^{2}-z^{2}=a^{2}w2x2y2z2=a2 (the equation of a four-dimensional hyperboloid).
Fig. 16.3 (a) A triangle on a 2sphere, whose angles sum to > π > π > pi>\pi>π. The geodesics of this space are great circles on the sphere. (b) A triangle on the Poincaré disc representation of hyperbolic space, whose angles add to < π < π < pi<\pi<π. The geodesics meet the boundary of the disc at right angles.
and the result is that the spacetime (as well as space) is flat, which is to say it has zero curvature. On the other hand, if k = 0 k = 0 k=0k=0k=0 and a ( t ) a ( t ) a(t)!=a(t) \neqa(t) const., then three-dimensional space is flat, but this flat space expands as a function of the time coordinate. In this case, some components of R R R\boldsymbol{R}R are non-zero and so spacetime is curved.

Example 16.4

The geometry to have in mind for the k = 0 k = 0 k=0k=0k=0 case is the flat, Euclidean plane, with all of its familiar features such as (i) parallel lines remaining parallel forever, and (ii) triangles with internal angles summing to π π pi\piπ.
Since, for our three-dimensional space, we have l ( t ) = a ( t ) ξ l ( t ) = a ( t ) ξ l(t)=a(t)xil(t)=a(t) \xil(t)=a(t)ξ then the ratio
(16.26) ( Proper circumference of circle proper radius ) = 2 π , (16.26)  Proper circumference of circle   proper radius  = 2 π , {:(16.26)((" Proper circumference of circle ")/(" proper radius "))=2pi",":}\begin{equation*} \left(\frac{\text { Proper circumference of circle }}{\text { proper radius }}\right)=2 \pi, \tag{16.26} \end{equation*}(16.26)( Proper circumference of circle  proper radius )=2π,
just as we have in flat two-dimensional space.
The case k = 1 k = 1 k=1k=1k=1
If k = 1 k = 1 k=1k=1k=1 then the line element is given by
(16.27) d s 2 = d t 2 + a ( t ) 2 [ d r 2 1 r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (16.27) d s 2 = d t 2 + a ( t ) 2 d r 2 1 r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.27)ds^(2)=-dt^(2)+a(t)^(2)[((d)r^(2))/(1-r^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{16.27} \end{equation*}(16.27)ds2=dt2+a(t)2[ dr21r2+r2( dθ2+sin2θ dϕ2)]
The curvature of k = 1 k = 1 k=1k=1k=1 space is positive. In Appendix D, we show how the spatial part of this metric can be embedded in four-dimensional Euclidean space as a sphere. 11 11 ^(11){ }^{11}11 It is shown with a dimension suppressed in Fig. 16.2(a).

Example 16.5

The k = 1 k = 1 k=1k=1k=1 geometry is sometime called elliptic or spherical. The example to have in mind is the two-dimensional surface of a sphere. Parallel lines in this geometry will eventually meet. A triangle in a two-dimensional space with elliptic geometry has internal angles that add up to greater than π π pi\piπ. In fact, we saw in Chapter 3 that if we draw a triangle on a 2-sphere of radius R R RRR [Fig. 16.3(a)] its area obeys Girard's theorem
(16.29) ( Area ) = R 2 ( α 1 + α 2 + α 3 π ) , (16.29) (  Area  ) = R 2 α 1 + α 2 + α 3 π , {:(16.29)(" Area ")=R^(2)(alpha_(1)+alpha_(2)+alpha_(3)-pi)",":}\begin{equation*} (\text { Area })=R^{2}\left(\alpha_{1}+\alpha_{2}+\alpha_{3}-\pi\right), \tag{16.29} \end{equation*}(16.29)( Area )=R2(α1+α2+α3π),
where α i α i alpha_(i)\alpha_{i}αi are the internal angles of the triangle.
Returning to three spatial dimensions and using the result that the proper circumference of a circle in the plane θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 is 2 π a ( t ) ξ 2 π a ( t ) ξ 2pi a(t)xi2 \pi a(t) \xi2πa(t)ξ, we have, on substituting for ( t ) ( t ) ℓ(t)\ell(t)(t) from eqn 16.20 ,
(16.30) ( Proper circumference of circle proper radius ) = 2 π a ( t ) ξ a ( t ) sin 1 ξ 2 π . (16.30)  Proper circumference of circle   proper radius  = 2 π a ( t ) ξ a ( t ) sin 1 ξ 2 π . {:(16.30)((" Proper circumference of circle ")/(" proper radius "))=(2pi a(t)xi)/(a(t)sin^(-1)xi) <= 2pi.:}\begin{equation*} \left(\frac{\text { Proper circumference of circle }}{\text { proper radius }}\right)=\frac{2 \pi a(t) \xi}{a(t) \sin ^{-1} \xi} \leq 2 \pi . \tag{16.30} \end{equation*}(16.30)( Proper circumference of circle  proper radius )=2πa(t)ξa(t)sin1ξ2π.
This is also the case if we take this ratio for a circle drawn on a 2 -sphere, just as we did in Chapter 3, justifying the intuition of our embedding argument.
The volume of the k = 1 k = 1 k=1k=1k=1 universe is given by
(16.31) V = χ = 0 π 4 π a ( t ) 3 sin 2 χ d χ = 2 π 2 a ( t ) 3 . (16.31) V = χ = 0 π 4 π a ( t ) 3 sin 2 χ d χ = 2 π 2 a ( t ) 3 . {:(16.31)V=int_(chi=0)^(pi)4pi a(t)^(3)sin^(2)chidchi=2pi^(2)a(t)^(3).:}\begin{equation*} V=\int_{\chi=0}^{\pi} 4 \pi a(t)^{3} \sin ^{2} \chi \mathrm{~d} \chi=2 \pi^{2} a(t)^{3} . \tag{16.31} \end{equation*}(16.31)V=χ=0π4πa(t)3sin2χ dχ=2π2a(t)3.
Since the space has this finite proper volume (owing to the limits on the variable χ χ chi\chiχ in eqn 16.17, which ranges from 0 χ π 0 χ π 0 <= chi <= pi0 \leq \chi \leq \pi0χπ ), it is said to be closed. However, what it does not have is a well-defined boundary, edge, or centre. This is similar to the situation for beings confined to the surface of a 2 -sphere. It's only when we embed the 2 -sphere in threedimensional space that we're able to characterize the ball as having a centre or boundary.
Fig. 16.4 (a) Radial light rays in a closed universe reach an observer at A, who sees a star at B in front of them and one at C behind them. (b) If we wait long enough, radial light can potentially traverse the whole universe.
12 12 ^(12){ }^{12}12 We show in Chapter 19 that in a physically realistic model there is not enough time for this to actually occur!
13 13 ^(13){ }^{13}13 The embedding required is
w = a cosh χ w = a cosh χ w=a cosh chiw=a \cosh \chiw=acoshχ,
x = a sinh χ sin θ cos ϕ x = a sinh χ sin θ cos ϕ x=a sinh chi sin theta cos phix=a \sinh \chi \sin \theta \cos \phix=asinhχsinθcosϕ,
y = a sinh χ sin θ sin ϕ y = a sinh χ sin θ sin ϕ y=a sinh chi sin theta sin phiy=a \sinh \chi \sin \theta \sin \phiy=asinhχsinθsinϕ,
z = a sinh χ cos θ z = a sinh χ cos θ z=a sinh chi cos thetaz=a \sinh \chi \cos \thetaz=asinhχcosθ.
The hyperboloid is defined by w 2 x 2 w 2 x 2 w^(2)-x^(2)-w^{2}-x^{2}-w2x2 y 2 z 2 = a 2 y 2 z 2 = a 2 y^(2)-z^(2)=a^(2)y^{2}-z^{2}=a^{2}y2z2=a2. Remember that it is possible to embed parts of the space in Euclidean space, but not all of it.
14 14 ^(14){ }^{14}14 Hyperbolic geometry is described in Penrose (2004) and Needham (1997). It was independent discovered by several people, but is sometimes called Lobachevskian geometry after Nikolai I. Lobachevsky (1792-1856). Lobachevsky is the subject of a song by satirist and mathematician Tom Lehrer, which was described as 'dazzlingly inventive in its shameless promotion of plagiarism', although Lehrer motion of plagiarism, although Lehrer
stressed the song was not meant as a slur, with Lobachevsky's name chosen 'solely for prosodic reasons'.
15 15 ^(15){ }^{15}15 Johann Heinrich Lambert (17281777)
16 16 ^(16){ }^{16}16 This is an example of a conformal representation. Conformal geometry is discussed in Chapter 19. Despite its name, the Poincaré disc was originally name, the Poincaré disc was originally
formulated by Eugenio Beltrami (1835formulated by Eugenio Beltrami (1835-
1900 ) in 1868 , but was rediscovered 1900) in 1868, but was rediscovered by Henri Poincaré fourteen years later. See Penrose (2004) for a short account of the history of hyperbolic space.
17 17 ^(17){ }^{17}17 We shall generally refer to hyperbolic ( k = 1 ) ( k = 1 ) (k=-1)(k=-1)(k=1), flat ( k = 0 ) ( k = 0 ) (k=0)(k=0)(k=0) and spherical ( k = 1 ) ( k = 1 ) (k=1)(k=1)(k=1) spaces below. They are some times respectively called open, flat and closed in the literature, although this is slightly ambiguous since flat space with k = 0 k = 0 k=0k=0k=0 is also open.
Fig. 16.5 Curves on a hyperboloid can be projected down on to the Poincare disc where the geodesics meet the boundary at right angles.
คThe second cause of redshift, due to light climbing out of a relativistic potential, is discussed further in Chapter 22 in the next part of the book.
It is notable that k = 1 k = 1 k=-1k=-1k=1 space does not have a finite proper volume: the integral for the volume diverges. This is because χ χ chi\chiχ ranges from 0 χ 0 χ 0 <= chi <= oo0 \leq \chi \leq \infty0χ in eqn 16.17 for k = 1 k = 1 k=-1k=-1k=1. As a result of its infinite volume, k = 1 k = 1 k=-1k=-1k=1 space is said to be open. (The k = 0 k = 0 k=0k=0k=0 space is also open.) 17 17 ^(17){ }^{17}17
Although useful for understanding intervals in the spacelike hypersurfaces, proper distances are not directly measurable quantities. In the next section, we turn to the key observable in cosmology: the redshift.

16.3 Redshift and cosmic expansion

In the last section, we saw how the evolution of a model universe is determined by the expansion factor a ( t ) a ( t ) a(t)a(t)a(t). The expansion of a universe is not something we can simply measure with a ruler. The problem, of course, is that a measuring device of a suitable size to measure the scales over which space expands will also be subject to the same expansion of space itself. In this section, we investigate the key observable in cosmology: a shift in frequency of light signals propagating over large distances that occurs due to changes in the geometry of spacetime. The shift is to lower frequencies for an expanding universe, and so is called a redshift. We have seen two reasons for redshift previously: the Doppler effect in special relativity, and the gravitational effect described in Chapter 6. Here we meet a third cause of redshift, the expansion of the universe itself. Such a redshift allows us to make measurements of the cosmological radius a ( t ) a ( t ) a(t)a(t)a(t).
An observer sits at the origin of our comoving coordinates at coordinate x μ = ( t ob , 0 , 0 , 0 ) x μ = t ob , 0 , 0 , 0 x^(mu)=(t_(ob),0,0,0)x^{\mu}=\left(t_{\mathrm{ob}}, 0,0,0\right)xμ=(tob,0,0,0), observing light signals sent from a distant galaxy at a (fixed) radial coordinate r = χ r = χ r=chir=\chir=χ, such that emission of photons occurs at coordinates ( t em , χ , 0 , 0 ) t em , χ , 0 , 0 (t_(em),chi,0,0)\left(t_{\mathrm{em}}, \chi, 0,0\right)(tem,χ,0,0). Light travels on null geodesics so that d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0 and, consequently, the crests of the light wave have an equation of motion given by
(16.36) d t 2 a ( t ) 2 d r 2 1 k r 2 = 0 (16.36) d t 2 a ( t ) 2 d r 2 1 k r 2 = 0 {:(16.36)dt^(2)-a(t)^(2)((d)r^(2))/(1-kr^(2))=0:}\begin{equation*} \mathrm{d} t^{2}-a(t)^{2} \frac{\mathrm{~d} r^{2}}{1-k r^{2}}=0 \tag{16.36} \end{equation*}(16.36)dt2a(t)2 dr21kr2=0
We examine the consequences of this in the following example.
Example 16.8
Rearranging and integrating along the null geodesic taken by the light ray, we have for a photon that 18 18 ^(18){ }^{18}18
(16.38) t em t ob d t a ( t ) = 0 χ d r ( 1 k r 2 ) 1 2 (16.38) t em t ob d t a ( t ) = 0 χ d r 1 k r 2 1 2 {:(16.38)int_(t_(em))^(t_(ob))((d)t)/(a(t))=int_(0)^(chi)(dr)/((1-kr^(2))^((1)/(2))):}\begin{equation*} \int_{t_{\mathrm{em}}}^{t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)}=\int_{0}^{\chi} \frac{\mathrm{d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}} \tag{16.38} \end{equation*}(16.38)temtob dta(t)=0χdr(1kr2)12
Now we imagine a second photon emitted at a time t em + δ t em t em + δ t em t_(em)+deltat_(em)t_{\mathrm{em}}+\delta t_{\mathrm{em}}tem+δtem and detected at t ob + δ t ob t ob + δ t ob t_(ob)+deltat_(ob)t_{\mathrm{ob}}+\delta t_{\mathrm{ob}}tob+δtob. As above, we have an equation that describes the null geodesic
(16.39) t em + δ t em t ob + δ t ob d t a ( t ) = 0 χ d r ( 1 k r 2 ) 1 2 (16.39) t em + δ t em t ob + δ t ob d t a ( t ) = 0 χ d r 1 k r 2 1 2 {:(16.39)int_(t_(em)+deltat_(em))^(t_(ob)+deltat_(ob))((d)t)/(a(t))=int_(0)^(chi)(dr)/((1-kr^(2))^((1)/(2))):}\begin{equation*} \int_{t_{\mathrm{em}}+\delta t_{\mathrm{em}}}^{t_{\mathrm{ob}}+\delta t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)}=\int_{0}^{\chi} \frac{\mathrm{d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}} \tag{16.39} \end{equation*}(16.39)tem+δtemtob+δtob dta(t)=0χdr(1kr2)12
The integral over the radial coordinate on the right-hand side of these two equations is identical (it should be: we're just describing two identical photon emission and detection events). Equating the left-hand side of the equations we have
(16.40) t em t ob d t a ( t ) = t em + δ t em t ob + δ t ob d t a ( t ) (16.40) t em t ob d t a ( t ) = t em + δ t em t ob + δ t ob d t a ( t ) {:(16.40)int_(t_(em))^(t_(ob))((d)t)/(a(t))=int_(t_(em)+deltat_(em))^(t_(ob)+deltat_(ob))((d)t)/(a(t)):}\begin{equation*} \int_{t_{\mathrm{em}}}^{t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)}=\int_{t_{\mathrm{em}}+\delta t_{\mathrm{em}}}^{t_{\mathrm{ob}}+\delta t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)} \tag{16.40} \end{equation*}(16.40)temtob dta(t)=tem+δtemtob+δtob dta(t)
Next, we use a trick: splitting the integrals up and summing. The idea is illustrated Next, we use a trick: splitting the ind
in Fig. 16.6. We write the left-hand side of eqn 16.40 as
(16.41) t em t ob d t a ( t ) = t em t em + δ t em d t a ( t ) + t em + δ t em t ob d t a ( t ) (16.41) t em t ob d t a ( t ) = t em t em + δ t em d t a ( t ) + t em + δ t em t ob d t a ( t ) {:(16.41)int_(t_(em))^(t_(ob))((d)t)/(a(t))=int_(t_(em))^(t_(em)+deltat_(em))((d)t)/(a(t))+int_(t_(em)+deltat_(em))^(t_(ob))((d)t)/(a(t)):}\begin{equation*} \int_{t_{\mathrm{em}}}^{t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)}=\int_{t_{\mathrm{em}}}^{t_{\mathrm{em}}+\delta t_{\mathrm{em}}} \frac{\mathrm{~d} t}{a(t)}+\int_{t_{\mathrm{em}}+\delta t_{\mathrm{em}}}^{t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)} \tag{16.41} \end{equation*}(16.41)temtob dta(t)=temtem+δtem dta(t)+tem+δtemtob dta(t)
Likewise, the integral on the right-hand side of eqn 16.40 can be written as
(16.42) t em + δ t em t ob + δ t ob d t a ( t ) = t em + δ t em t ob d t a ( t ) + t ob t ob + δ t ob d t a ( t ) (16.42) t em + δ t em t ob + δ t ob d t a ( t ) = t em + δ t em t ob d t a ( t ) + t ob t ob + δ t ob d t a ( t ) {:(16.42)int_(t_(em)+deltat_(em))^(t_(ob)+deltat_(ob))((d)t)/(a(t))=int_(t_(em)+deltat_(em))^(t_(ob))((d)t)/(a(t))+int_(t_(ob))^(t_(ob)+deltat_(ob))((d)t)/(a(t)):}\begin{equation*} \int_{t_{\mathrm{em}}+\delta t_{\mathrm{em}}}^{t_{\mathrm{ob}}+\delta t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)}=\int_{t_{\mathrm{em}}+\delta t_{\mathrm{em}}}^{t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)}+\int_{t_{\mathrm{ob}}}^{t_{\mathrm{ob}}+\delta t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)} \tag{16.42} \end{equation*}(16.42)tem+δtemtob+δtob dta(t)=tem+δtemtob dta(t)+tobtob+δtob dta(t)
This sleight of hand allows us to cancel out the identical parts [i.e. the limits represent the long period between the emission of the second photon and the observation of the first one, which is common to both sides of the equation, as shown in Fig. 16.6]. This leaves
(16.43) t em t em + δ t em d t a ( t ) = t ob t ob + δ t ob d t a ( t ) (16.43) t em t em + δ t em d t a ( t ) = t ob t ob + δ t ob d t a ( t ) {:(16.43)int_(t_(em))^(t_(em)+deltat_(em))((d)t)/(a(t))=int_(t_(ob))^(t_(ob)+deltat_(ob))((d)t)/(a(t)):}\begin{equation*} \int_{t_{\mathrm{em}}}^{t_{\mathrm{em}}+\delta t_{\mathrm{em}}} \frac{\mathrm{~d} t}{a(t)}=\int_{t_{\mathrm{ob}}}^{t_{\mathrm{ob}}+\delta t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)} \tag{16.43} \end{equation*}(16.43)temtem+δtem dta(t)=tobtob+δtob dta(t)
which, in words, tells us that the comoving distance covered by the first photon before the emission of the second, is the same as the comoving distance covered by the second photon after the detection of the first. The time between emission (or detection) of the two light photons is assumed very small on cosmological scales, and so each of the integrands is very nearly constant over the limits. We can therefore replace the integrals with the approximate expression
(16.44) δ t em a ( t em ) = δ t ob a ( t ob ) (16.44) δ t em a t em = δ t ob a t ob {:(16.44)(deltat_(em))/(a(t_(em)))=(deltat_(ob))/(a(t_(ob))):}\begin{equation*} \frac{\delta t_{\mathrm{em}}}{a\left(t_{\mathrm{em}}\right)}=\frac{\delta t_{\mathrm{ob}}}{a\left(t_{\mathrm{ob}}\right)} \tag{16.44} \end{equation*}(16.44)δtema(tem)=δtoba(tob)
This provides us with a relationship between intervals at the observer and emitter of the signal and the expansion factor evaluated at their respective coordinate times.
Rearranging the previous expression we have
(16.45) δ t em δ t ob = a ( t em ) a ( t ob ) (16.45) δ t em δ t ob = a t em a t ob {:(16.45)(deltat_(em))/(deltat_(ob))=(a(t_(em)))/(a(t_(ob))):}\begin{equation*} \frac{\delta t_{\mathrm{em}}}{\delta t_{\mathrm{ob}}}=\frac{a\left(t_{\mathrm{em}}\right)}{a\left(t_{\mathrm{ob}}\right)} \tag{16.45} \end{equation*}(16.45)δtemδtob=a(tem)a(tob)
If δ t δ t delta t\delta tδt represents a proper period for oscillations of the light (i.e. the crests of a light wave, separated in emission time by δ t em δ t em deltat_(em)\delta t_{\mathrm{em}}δtem and in observation time by δ t ob δ t ob deltat_(ob)\delta t_{\mathrm{ob}}δtob ), we can convert this ratio to a ratio of frequencies and write
(16.46) ω ob ω em = a ( t em ) a ( t ob ) (16.46) ω ob ω em = a t em a t ob {:(16.46)(omega_(ob))/(omega_(em))=(a(t_(em)))/(a(t_(ob))):}\begin{equation*} \frac{\omega_{\mathrm{ob}}}{\omega_{\mathrm{em}}}=\frac{a\left(t_{\mathrm{em}}\right)}{a\left(t_{\mathrm{ob}}\right)} \tag{16.46} \end{equation*}(16.46)ωobωem=a(tem)a(tob)
Notice that the shift depends only on the radius function a ( t ) a ( t ) a(t)a(t)a(t) evaluated at the time of emission and of observation.

Example 16.9

For an expanding universe, we have a ( t ob ) > a ( t em ) a t ob > a t em a(t_(ob)) > a(t_(em))a\left(t_{\mathrm{ob}}\right)>a\left(t_{\mathrm{em}}\right)a(tob)>a(tem) and so ω ob < ω em ω ob < ω em omega_(ob) < omega_(em)\omega_{\mathrm{ob}}<\omega_{\mathrm{em}}ωob<ωem. Each point is receding away from every other point and light signals are redshifted. Conversely, for a contracting universe, light signals will be blueshifted. Referring back to our closed universe in Fig. 16.4(a), if a ( t ) a ( t ) a(t)a(t)a(t) increases with time, the light received via the route shown between the stars at B and C and the observer at A is redshifted. If a ( t ) a ( t ) a(t)a(t)a(t) decreases with time, these light rays will be blueshifted.
Fig. 16.6 The derivation of the cosmological redshift relies on equating the comoving distance d t / a ( t ) d t / a ( t ) intdt//a(t)\int \mathrm{d} t / a(t)dt/a(t) for a ray emitted at t em t em t_(em)t_{\mathrm{em}}tem and observed at t ob t ob t_(ob)t_{\mathrm{ob}}tob to one emitted at t em + δ t em t em + δ t em t_(em)+deltat_(em)t_{\mathrm{em}}+\delta t_{\mathrm{em}}tem+δtem and observed at t ob + δ t em t ob + δ t em t_(ob)+deltat_(em)t_{\mathrm{ob}}+\delta t_{\mathrm{em}}tob+δtem. The part between t em + δ t em t em + δ t em t_(em)+deltat_(em)t_{\mathrm{em}}+\delta t_{\mathrm{em}}tem+δtem and t ob t ob t_(ob)t_{\mathrm{ob}}tob (dashed) is comment to both sides of the equation, and therefore cancels.
19 19 ^(19){ }^{19}19 It turns out that this property, λ a λ a lambda prop a\lambda \propto aλa works not only for photons, but also for neutrinos and gravitational waves.
We have found that the frequency of the photon is inversely proportional to the scale factor, meaning that the wavelength of the photon is proportional to the scale factor. Another way of thinking of the cosmological redshift is that it is as if the wavelength of a photon is frozen into the Universe and stretches with it. 19 19 ^(19){ }^{19}19 Conventionally, redshifts are written in terms of a redshift parameter z z zzz defined in terms of the wavelength at emission and observation as
(16.47) z = λ ob λ em λ em (16.47) z = λ ob λ em λ em {:(16.47)z=(lambda_(ob)-lambda_(em))/(lambda_(em)):}\begin{equation*} z=\frac{\lambda_{\mathrm{ob}}-\lambda_{\mathrm{em}}}{\lambda_{\mathrm{em}}} \tag{16.47} \end{equation*}(16.47)z=λobλemλem
Very often, we will take the scale factor at the time of observation to be unity, so this simplifies to
(16.48) 1 + z = 1 a ( t em ) (16.48) 1 + z = 1 a t em {:(16.48)1+z=(1)/(a(t_(em))):}\begin{equation*} 1+z=\frac{1}{a\left(t_{\mathrm{em}}\right)} \tag{16.48} \end{equation*}(16.48)1+z=1a(tem)
1 + z = λ ob λ em = a ( t ob ) a ( t em ) 1 + z = λ ob λ em = a t ob a t em 1+z=(lambda_(ob))/(lambda_(em))=(a(t_(ob)))/(a(t_(em)))1+z=\frac{\lambda_{\mathrm{ob}}}{\lambda_{\mathrm{em}}}=\frac{a\left(t_{\mathrm{ob}}\right)}{a\left(t_{\mathrm{em}}\right)}1+z=λobλem=a(tob)a(tem)
We conclude that a measurement of z z zzz gives us access to information about the expansion factor a ( t ) a ( t ) a(t)a(t)a(t). Note that when discussing expanding model universes, we often define the zero of time such that light emitted then has infinite redshift z [ z [ z[z[z[ i.e. a ( t = 0 ) = 0 ] a ( t = 0 ) = 0 ] a(t=0)=0]a(t=0)=0]a(t=0)=0].

16.4 The initial singularity

Robertson-Walker spacetimes can be used to describe expanding universes that can be spherical, flat or hyperbolic. In the next chapter, we look at the consequences of filling these spacetimes with homogeneous, isotropic distributions of matter. This results in universes whose densities decrease as time runs forward. The natural question arises about what happens if we run time backwards, with the density of the universe increasing. We shall find that for most of our models, the factor a ( t ) a ( t ) a(t)a(t)a(t) is zero at some time t 0 t 0 t_(0)t_{0}t0, a finite length of time ago. The density of the universe increases without bound as t 0 t 0 t_(0)t_{0}t0 is approached, and this results in a singularity in the physical fields at time t 0 t 0 t_(0)t_{0}t0, which can reasonably be assigned to be the time of the beginning of the universe. This is, of course, the Big Bang. The singularity is even worse than it would appear in a Newtonian model of this physics, where the density of the universe becomes infinite. Since the metric field is the key field in general relativity, a singularity in the density of matter implies a singularity in the very fabric of space and time, which amounts to an infinite curvature. This is something with which our classical field theory of gravitation cannot cope, and so we are forced to cut this point out of our description of spacetime. No known physical laws can apply at this initial singularity.
The initial singularity in spacetime is one of the most interesting things about the Robertson-Walker spacetimes. We shall not dwell on it in the next chapters, where we solve the Einstein equation for several distributions of matter filling these spaces. We return to the singularities in Chapter 19, where we shall find a way for treating the singularity as an effective edge of spacetime.

Chapter summary

  • There are three Robertson-Walker geometries described by the metric
d s 2 = d t 2 + a ( t ) 2 [ d r 2 1 k r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] d s 2 = d t 2 + a ( t ) 2 d r 2 1 k r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-dt^(2)+a(t)^(2)[((d)r^(2))/(1-kr^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]\mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-k r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right]ds2=dt2+a(t)2[ dr21kr2+r2( dθ2+sin2θ dϕ2)]
They are all isotropic and homogeneous and are classified as spherical ( k = 1 ) ( k = 1 ) (k=1)(k=1)(k=1), flat ( k = 0 ) ( k = 0 ) (k=0)(k=0)(k=0) and hyperbolic ( k = 1 ) ( k = 1 ) (k=-1)(k=-1)(k=1). They evolve as a function of time.
  • The evolution of the universes, given in terms of the effective radius a ( t ) a ( t ) a(t)a(t)a(t), which is related to the measured redshift z z zzz via
(16.50) 1 + z = a ( t ob ) a ( t em ) . (16.50) 1 + z = a t ob a t em . {:(16.50)1+z=(a(t_(ob)))/(a(t_(em))).:}\begin{equation*} 1+z=\frac{a\left(t_{\mathrm{ob}}\right)}{a\left(t_{\mathrm{em}}\right)} . \tag{16.50} \end{equation*}(16.50)1+z=a(tob)a(tem).
  • The Robertson-Walker universes have, at their start, the potential for an initial Big-Bang singularity.

Exercises

(16.1) Verify that ( 3 ) R i j k l = K ( γ i k γ j l γ i l γ j k ) ( 3 ) R i j k l = K γ i k γ j l γ i l γ j k ^((3))R_(ijkl)=K(gamma_(ik)gamma_(jl)-gamma_(il)gamma_(jk)){ }^{(3)} R_{i j k l}=K\left(\gamma_{i k} \gamma_{j l}-\gamma_{i l} \gamma_{j k}\right)(3)Rijkl=K(γikγjlγilγjk) has the expected symmetries of the Riemann tensor
(16.2) Verify eqn 16.10
(16.3) Compute the connection coefficients for the Robertson-Walker spacetimes and verify they are those given in this chapter.
(16.4) Starting with d σ 2 = γ i j d x i d x j d σ 2 = γ i j d x i d x j dsigma^(2)=gamma_(ij)dx^(i)dx^(j)\mathrm{d} \sigma^{2}=\gamma_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j}dσ2=γij dxi dxj and transforming to spherical coordinates results in the general three-dimensional metric
(16.51) d σ 2 = e 2 Λ ( r ) d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (16.51) d σ 2 = e 2 Λ ( r ) d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(16.51)dsigma^(2)=e^(2Lambda(r))dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} \sigma^{2}=\mathrm{e}^{2 \Lambda(r)} \mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{16.51} \end{equation*}(16.51)dσ2=e2Λ(r)dr2+r2( dθ2+sin2θ dϕ2)
where e 2 Λ ( r ) e 2 Λ ( r ) e^(2Lambda(r))\mathrm{e}^{2 \Lambda(r)}e2Λ(r) is some function of r r rrr that we shall determine.
(a) Identify vielbein that transform from the coordinate frame to the orthonormal frame.
(b) Comparing with the results from Exercise 11.1 in Chapter 11, show that the components of the Riemann tensor in the orthonormal frame can be written as
R θ ^ ϕ ^ θ ^ ϕ ˙ = F , (16.52) R ϕ ^ θ ^ r ^ ϕ ^ = F ¯ , R r ^ θ ^ = r ¯ θ ^ R θ ^ ϕ ^ θ ^ ϕ ˙ = F , (16.52) R ϕ ^ θ ^ r ^ ϕ ^ = F ¯ , R r ^ θ ^ = r ¯ θ ^ {:[R^( hat(theta) hat(phi))_( hat(theta)phi^(˙))=F","],[(16.52)R^( hat(phi) hat(theta))^( hat(r) hat(phi))=- bar(F)","],[R^( hat(r) hat(theta))= bar(r) hat(theta)]:}\begin{gather*} R^{\hat{\theta} \hat{\phi}}{ }_{\hat{\theta} \dot{\phi}}=F, \\ R^{\hat{\phi} \hat{\theta}}{ }^{\hat{r} \hat{\phi}}=-\bar{F}, \tag{16.52}\\ R^{\hat{r} \hat{\theta}}=\bar{r} \hat{\theta} \end{gather*}Rθ^ϕ^θ^ϕ˙=F,(16.52)Rϕ^θ^r^ϕ^=F¯,Rr^θ^=r¯θ^
where
F = 1 r 2 ( 1 e 2 Λ ) , (16.53) F ¯ = 1 r e 2 Λ Λ F = 1 r 2 1 e 2 Λ , (16.53) F ¯ = 1 r e 2 Λ Λ {:[F=(1)/(r^(2))*(1-e^(-2Lambda))","],[(16.53) bar(F)=(1)/(r)*e^(-2Lambda)Lambda^(')]:}\begin{align*} F & =\frac{1}{r^{2}} \cdot\left(1-\mathrm{e}^{-2 \Lambda}\right), \\ \bar{F} & =\frac{1}{r} \cdot \mathrm{e}^{-2 \Lambda} \Lambda^{\prime} \tag{16.53} \end{align*}F=1r2(1e2Λ),(16.53)F¯=1re2ΛΛ
(c) What are the components of the Ricci tensor in the orthonormal frame? Show that in the coordinate frame they are
(16.54) R r r = Λ r , R θ θ = R ϕ ϕ sin 2 θ = ( 1 e 2 Λ ) + r Λ e 2 Λ (16.54) R r r = Λ r , R θ θ = R ϕ ϕ sin 2 θ = 1 e 2 Λ + r Λ e 2 Λ {:(16.54)R_(rr)=(Lambda^('))/(r)","quadR_(theta theta)=(R_(phi phi))/(sin^(2)theta)=(1-e^(-2Lambda))+rLambda^(')e^(-2Lambda):}\begin{equation*} R_{r r}=\frac{\Lambda^{\prime}}{r}, \quad R_{\theta \theta}=\frac{R_{\phi \phi}}{\sin ^{2} \theta}=\left(1-\mathrm{e}^{-2 \Lambda}\right)+r \Lambda^{\prime} \mathrm{e}^{-2 \Lambda} \tag{16.54} \end{equation*}(16.54)Rrr=Λr,Rθθ=Rϕϕsin2θ=(1e2Λ)+rΛe2Λ
(d) Show that for a space of constant curvature, where R i j = 2 K g i j R i j = 2 K g i j R_(ij)=2Kg_(ij)R_{i j}=2 K g_{i j}Rij=2Kgij, we have
(16.55) Λ = 2 K r e 2 Λ , ( 1 e 2 Λ ) + r Λ e Λ = 2 K r 2 (16.55) Λ = 2 K r e 2 Λ , 1 e 2 Λ + r Λ e Λ = 2 K r 2 {:(16.55)Lambda^(')=2Kre^(2Lambda)","quad(1-e^(-2Lambda))+rLambda^(')e^(-Lambda)=2Kr^(2):}\begin{equation*} \Lambda^{\prime}=2 K r \mathrm{e}^{2 \Lambda}, \quad\left(1-\mathrm{e}^{-2 \Lambda}\right)+r \Lambda^{\prime} \mathrm{e}^{-\Lambda}=2 K r^{2} \tag{16.55} \end{equation*}(16.55)Λ=2Kre2Λ,(1e2Λ)+rΛeΛ=2Kr2
Hint: It's easiest to work in the orthonormal frame.
(e) By solving the equations given in (d), show that we must have a line element
d σ 2 = d r 2 1 K r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d σ 2 = d r 2 1 K r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 dsigma^(2)=(dr^(2))/(1-Kr^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))\mathrm{d} \sigma^{2}=\frac{\mathrm{d} r^{2}}{1-K r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)dσ2=dr21Kr2+r2( dθ2+sin2θ dϕ2)
(16.5) Using the arguments from this chapter, we can say that if the curvature of a spacetime is isotropic, the
components of the Riemann tensor can always be written in the form
(16.57) R μ ν α β = C ( g μ α g ν β g μ β g ν α ) (16.57) R μ ν α β = C g μ α g ν β g μ β g ν α {:(16.57)R_(mu nu alpha beta)=C(g_(mu alpha)g_(nu beta)-g_(mu beta)g_(nu alpha)):}\begin{equation*} R_{\mu \nu \alpha \beta}=C\left(g_{\mu \alpha} g_{\nu \beta}-g_{\mu \beta} g_{\nu \alpha}\right) \tag{16.57} \end{equation*}(16.57)Rμναβ=C(gμαgνβgμβgνα)
Use the Bianchi identity (from Chapter 13), along with the compatibility condition that g μ ν ; α = 0 g μ ν ; α = 0 g_(mu nu;alpha)=0g_{\mu \nu ; \alpha}=0gμν;α=0 (from Chapter 7), to show that C C CCC must be a constant.
(16.6) In the last chapter, we described (3+1)dimensional spaces of constant curvature as having R μ ν = 1 4 R g μ ν R μ ν = 1 4 R g μ ν R_(mu nu)=(1)/(4)Rg_(mu nu)R_{\mu \nu}=\frac{1}{4} R g_{\mu \nu}Rμν=14Rgμν. Show this is the case by contracting the expression in the previous problem. Generalize the result for d d ddd-dimensional spacetime.
(16.7) What is the three-dimensional Ricci scalar ( 3 ) R ( 3 ) R ^((3))R{ }^{(3)} R(3)R for the spatial part of the Robertson-Walker spacetime, written in terms of a ( t ) a ( t ) a(t)a(t)a(t) ? How can this be used to write ( 3 ) R i j ( 3 ) R i j ^((3))R_(ij){ }^{(3)} R_{i j}(3)Rij in terms of γ i j γ i j gamma_(ij)\gamma_{i j}γij ?
(16.8) A galaxy with redshift z z zzz and is observed to be at a proper distance ob ob ℓ_(ob)\ell_{\mathrm{ob}}ob away from us. What was the proper distance em em ℓ_(em)\ell_{\mathrm{em}}em to the galaxy at the time when its light was emitted?
(16.9) Show that if we differentiate 1 + z = a ( t ob ) / a ( t em ) 1 + z = a t ob / a t em 1+z=a(t_(ob))//a(t_(em))1+z=a\left(t_{\mathrm{ob}}\right) / a\left(t_{\mathrm{em}}\right)1+z=a(tob)/a(tem) with respect to the time at the observer t ob t ob t_(ob)t_{\mathrm{ob}}tob, we find that the rate of expansion of the Universe at the time of light emission is given by
(16.58) H ( t em ) = H ( t ob ) ( 1 + z ) z t ob (16.58) H t em = H t ob ( 1 + z ) z t ob {:(16.58)H(t_(em))=H(t_(ob))(1+z)-(del z)/(delt_(ob)):}\begin{equation*} H\left(t_{\mathrm{em}}\right)=H\left(t_{\mathrm{ob}}\right)(1+z)-\frac{\partial z}{\partial t_{\mathrm{ob}}} \tag{16.58} \end{equation*}(16.58)H(tem)=H(tob)(1+z)ztob

The Friedmann equations

Now hast thou but one bare hour to live, And then thou must be damn'd perpetually! Stand still, you ever-moving spheres of heaven, That time may cease, and midnight never come. Christopher Marlowe (1564-1593) Doctor Faustus
Margaret Fuller: I accept the Universe
Thomas Carlyle: Gad! she'd better!
Our mission remains to determine the origin and fate of model universes using the Einstein equation. With the geometry of the RobertsonWalker spaces supplied by the metric fields of the last chapter, we have access to the left-hand side of Einstein's equation. Now we need to fill in the right-hand side of the equation by specifying the matter content of the universes. This is our task in this chapter. Filling model universes with different idealized distributions of matter will lead us to two (equivalent) equations of motion for the expansion factor a ( t ) a ( t ) a(t)a(t)a(t), known as the Friedmann equations. These allow us access to a wealth of different models with which to amuse ourselves in the next chapter.
Example 17.1
Before ploughing into the general-relativistic version, here is a simple argument that gives some physical insight, albeit working things out in the Newtonian limit. The total energy U = T + V U = T + V U=T+VU=T+VU=T+V of a particle at radius r r rrr from the centre of a expanding homogeneous medium of density ρ ρ rho\rhoρ is the sum of kinetic energy T = 1 2 m r ˙ 2 T = 1 2 m r ˙ 2 T=(1)/(2)mr^(˙)^(2)T=\frac{1}{2} m \dot{r}^{2}T=12mr˙2 and potential energy U = G M m / r U = G M m / r U=-GMm//rU=-G M m / rU=GMm/r. Writing r = a ( t ) x r = a ( t ) x vec(r)=a(t) vec(x)\vec{r}=a(t) \vec{x}r=a(t)x, where x x vec(x)\vec{x}x is a comoving coordinate, and M = 4 3 π r 3 ρ M = 4 3 π r 3 ρ M=(4)/(3)pir^(3)rhoM=\frac{4}{3} \pi r^{3} \rhoM=43πr3ρ, we have
(17.1) U = 1 2 m a ˙ 2 x 2 4 π G 3 ρ a 2 x 2 m (17.1) U = 1 2 m a ˙ 2 x 2 4 π G 3 ρ a 2 x 2 m {:(17.1)U=(1)/(2)ma^(˙)^(2)x^(2)-(4pi G)/(3)rhoa^(2)x^(2)m:}\begin{equation*} U=\frac{1}{2} m \dot{a}^{2} x^{2}-\frac{4 \pi G}{3} \rho a^{2} x^{2} m \tag{17.1} \end{equation*}(17.1)U=12ma˙2x24πG3ρa2x2m
which rearranges to
(17.2) ( a ˙ a ) 2 + k a 2 = 8 π G ρ 3 (17.2) a ˙ a 2 + k a 2 = 8 π G ρ 3 {:(17.2)(((a^(˙)))/(a))^(2)+(k)/(a^(2))=(8pi G rho)/(3):}\begin{equation*} \left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=\frac{8 \pi G \rho}{3} \tag{17.2} \end{equation*}(17.2)(a˙a)2+ka2=8πGρ3
where k = 2 U m x 2 k = 2 U m x 2 k=-(2U)/(mx^(2))k=-\frac{2 U}{m x^{2}}k=2Umx2 is a time-independent constant. Remarkably, eqn 17.2 is essentially the first Friedmann equation (that we shall shortly derive), and so one can already appreciate that it arises from energy considerations. But after this warm-up exercise, it's time to derive the Friedmann equations using the general-relativistic techniques that we ve built up. qquad\qquad

17.1 Enter energy-momentum
17.2 Enter thermodynamics 183 17.3 Dust and radiation 184 Chapter summary 187 Exercises
1 1 ^(1){ }^{1}1 Alexander Friedmann (1888-1925). Friedmann died tragically young from typhoid. Legend has it that this was brought on by eating an unwashed pear that he had bought at a station on the way back from his honeymoon. The conventional spelling of Friedmann's surname is apparently due to a mistake of Einstein's. In a note Einstein published in 1923, Einstein referred to him as 'Friedmann', although he had signed himself 'Friedman' in his own paper of 1922. According to Ari Belenkiy: 'It seemed Friedman followed the "advice" and submitted his 1924 paper with two n's.' (See A. Belenkiy, arXiv:1302.1498.)
2 2 ^(2){ }^{2}2 We saw the same components for the energy-momentum tensor in the orthonormal frame in Chapter 10 for the k = 0 k = 0 k=0k=0k=0 case. See the Exercise 17.1 for verification that they're appropriate for k = ± 1 k = ± 1 k=+-1k= \pm 1k=±1.
3 3 ^(3){ }^{3}3 The Robertson-Walker line element is
d s 2 = d t 2 + a ( t ) 2 [ d r 2 1 k r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] d s 2 = d t 2 + a ( t ) 2 d r 2 1 k r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-dt^(2)+a(t)^(2)[((d)r^(2))/(1-kr^(2)):}],[{:+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]]:}\begin{aligned} \mathrm{d} s^{2}= & -\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-k r^{2}}\right. \\ & \left.+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \end{aligned}ds2=dt2+a(t)2[ dr21kr2+r2( dθ2+sin2θ dϕ2)]
We have metric components
g t t = 1 , g r r = a 2 11 k r 2 , g θ θ = a 2 r 2 , g ϕ ϕ = a 2 r 2 sin 2 θ . g t t = 1 , g r r = a 2 11 k r 2 , g θ θ = a 2 r 2 , g ϕ ϕ = a 2 r 2 sin 2 θ . {:[g_(tt)=-1","quadg_(rr)=(a^(2))/(11-kr^(2))","],[g_(theta theta)=a^(2)r^(2)","quadg_(phi phi)=a^(2)r^(2)sin^(2)theta.]:}\begin{gathered} g_{t t}=-1, \quad g_{r r}=\frac{a^{2}}{11-k r^{2}}, \\ g_{\theta \theta}=a^{2} r^{2}, \quad g_{\phi \phi}=a^{2} r^{2} \sin ^{2} \theta . \end{gathered}gtt=1,grr=a211kr2,gθθ=a2r2,gϕϕ=a2r2sin2θ.
e vielbein components are
( e t ) t ^ = 1 , ( e r ) r ^ = a ( 1 k r 2 ) 1 2 , ( e θ ) θ ^ = a r , ( e ϕ ) ϕ ^ = a r sin θ . e t t ^ = 1 , e r r ^ = a 1 k r 2 1 2 , e θ θ ^ = a r , e ϕ ϕ ^ = a r sin θ . {:[(e_(t))^( hat(t))=1","(e_(r))^( hat(r))=(a)/((1-kr^(2))^((1)/(2)))","],[(e_(theta))^( hat(theta))=ar","(e_(phi))^( hat(phi))=ar sin theta.]:}\begin{aligned} \left(\boldsymbol{e}_{t}\right)^{\hat{t}} & =1, & \left(\boldsymbol{e}_{r}\right)^{\hat{r}}=\frac{a}{\left(1-k r^{2}\right)^{\frac{1}{2}}}, \\ \left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}} & =a r, & \left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}=a r \sin \theta . \end{aligned}(et)t^=1,(er)r^=a(1kr2)12,(eθ)θ^=ar,(eϕ)ϕ^=arsinθ.
The Ricci tensor has non-zero components in the orthonormal frame of
R i t t ^ = 3 a ¨ a R i i ` ^ = a ¨ a + 2 ( a ˙ a ) 2 + 2 k a . R i t t ^ = 3 a ¨ a R i i ` ^ = a ¨ a + 2 a ˙ a 2 + 2 k a . {:[R_(it hat(t))=-3((a^(¨)))/(a)],[R_( hat(i(i^(`))))=((a^(¨)))/(a)+2(((a^(˙)))/(a))^(2)+2(k)/(a).]:}\begin{aligned} & R_{i t \hat{t}}=-3 \frac{\ddot{a}}{a} \\ & R_{\hat{i \grave{i}}}=\frac{\ddot{a}}{a}+2\left(\frac{\dot{a}}{a}\right)^{2}+2 \frac{k}{a} . \end{aligned}Ritt^=3a¨aRii`^=a¨a+2(a˙a)2+2ka.
The Ricci scalar is
R = 6 ( a ¨ a + a ˙ 2 a 2 + k a 2 ) . R = 6 a ¨ a + a ˙ 2 a 2 + k a 2 . R=6(((a^(¨)))/(a)+(a^(˙)^(2))/(a^(2))+(k)/(a^(2))).R=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}+\frac{k}{a^{2}}\right) .R=6(a¨a+a˙2a2+ka2).
The vielbein suggests that the spacelike components of the energy-momentum tensor can be thought of as comprising time-dependent effective pressures p ( t ) p ( t ) p(t)p(t)p(t) time-dependent effective pressures p ( t ) p ( t ) p(t)p(t)p(t)
proportional to 1 / a ( t ) 2 1 / a ( t ) 2 1//a(t)^(2)1 / a(t)^{2}1/a(t)2. This makes proportional to 1 / a ( t ) 2 1 / a ( t ) 2 1//a(t)^(2)1 / a(t)^{2}1/a(t)2. This makes
sense as the expansion of space caused sense as the expansion of space caused
by a ( t ) a ( t ) a(t)a(t)a(t) should lead to a reduction in by a ( t ) a ( t ) a(t)a(t)a(t) should lead to a reduction in
pressure (i.e. the force per unit area) in the fluid.

17.1 Enter energy-momentum

Following the pioneering work of Alexander Friedmann, 1 1 ^(1){ }^{1}1 we shall fill the Robertson-Walker spaces with both vacuum energy (with energy density Λ Λ Lambda\LambdaΛ ) and a perfect fluid (i.e. a fluid without any dissipation). Written in spherical polars in the orthonormal frame, the energy-momentum tensor for a perfect fluid is 2 2 ^(2){ }^{2}2
(17.3) T t ^ t ^ = ρ , T r ^ r ^ = T θ ^ θ ^ = T ϕ ^ ϕ ^ = p , (17.3) T t ^ t ^ = ρ , T r ^ r ^ = T θ ^ θ ^ = T ϕ ^ ϕ ^ = p , {:(17.3)T_( hat(t) hat(t))=rho","quadT_( hat(r) hat(r))=T_( hat(theta) hat(theta))=T_( hat(phi) hat(phi))=p",":}\begin{equation*} T_{\hat{t} \hat{t}}=\rho, \quad T_{\hat{r} \hat{r}}=T_{\hat{\theta} \hat{\theta}}=T_{\hat{\phi} \hat{\phi}}=p, \tag{17.3} \end{equation*}(17.3)Tt^t^=ρ,Tr^r^=Tθ^θ^=Tϕ^ϕ^=p,
where ρ ρ rho\rhoρ is the mass-energy density and p p ppp is the pressure. Armed with this, and the Ricci tensor for the Robertson-Walker spaces from the previous chapter, we can write down the components of the Einstein equation that governs the kinematics of these spaces. This results in two famous equations of motion known as the Friedmann equations.

Example 17.2

We work in the orthonormal frame with basis vectors e ı ^ , e r ^ , e θ ^ , e ϕ ^ e ı ^ , e r ^ , e θ ^ , e ϕ ^ e_( hat(ı)),e_( hat(r)),e_( hat(theta)),e_( hat(phi))\boldsymbol{e}_{\hat{\imath}}, \boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}, \boldsymbol{e}_{\hat{\phi}}eı^,er^,eθ^,eϕ^. Equating the 0̂0̂th components of the Einstein equation 3 3 ^(3){ }^{3}3 we have
R 0 ^ 0 ^ g 0 ^ 0 ^ 2 R = 8 π T 0 ^ 0 ^ Λ g 0 ^ 0 ^ R 0 ^ 0 ^ g 0 ^ 0 ^ 2 R = 8 π T 0 ^ 0 ^ Λ g 0 ^ 0 ^ R_( hat(0) hat(0))-(g_( hat(0) hat(0)))/(2)R=8piT_( hat(0) hat(0))-Lambdag_( hat(0) hat(0))R_{\hat{0} \hat{0}}-\frac{g_{\hat{0} \hat{0}}}{2} R=8 \pi T_{\hat{0} \hat{0}}-\Lambda g_{\hat{0} \hat{0}}R0^0^g0^0^2R=8πT0^0^Λg0^0^
3 a ¨ a + 3 [ a ¨ a + ( a ˙ a ) 2 + k a 2 ] = 8 π ρ + Λ 3 a ¨ a + 3 a ¨ a + a ˙ a 2 + k a 2 = 8 π ρ + Λ -3((a^(¨)))/(a)+3[((a^(¨)))/(a)+(((a^(˙)))/(a))^(2)+(k)/(a^(2))]=8pi rho+Lambda-3 \frac{\ddot{a}}{a}+3\left[\frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}\right]=8 \pi \rho+\Lambda3a¨a+3[a¨a+(a˙a)2+ka2]=8πρ+Λ
(17.4) ( a ˙ a ) 2 + k a 2 = 8 π ρ 3 + Λ 3 (17.4) a ˙ a 2 + k a 2 = 8 π ρ 3 + Λ 3 {:(17.4)(((a^(˙)))/(a))^(2)+(k)/(a^(2))=(8pi rho)/(3)+(Lambda)/(3):}\begin{equation*} \left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=\frac{8 \pi \rho}{3}+\frac{\Lambda}{3} \tag{17.4} \end{equation*}(17.4)(a˙a)2+ka2=8πρ3+Λ3
Next, we equate the i ^ i ^ i ^ i ^ hat(i) hat(i)\hat{i} \hat{i}i^i^ th components and find
[ a ¨ a + 2 ( a ˙ a ) 2 + 2 k a ] 3 [ a ¨ a + ( a ˙ a ) 2 + k a 2 ] = 8 π p Λ (17.5) 2 a ¨ a + ( a ˙ a ) 2 + k a 2 = 8 π p + Λ . a ¨ a + 2 a ˙ a 2 + 2 k a 3 a ¨ a + a ˙ a 2 + k a 2 = 8 π p Λ (17.5) 2 a ¨ a + a ˙ a 2 + k a 2 = 8 π p + Λ . {:[[((a^(¨)))/(a)+2(((a^(˙)))/(a))^(2)+2(k)/(a)]-3[((a^(¨)))/(a)+(((a^(˙)))/(a))^(2)+(k)/(a^(2))]=8pi p-Lambda],[(17.5)2((a^(¨)))/(a)+(((a^(˙)))/(a))^(2)+(k)/(a^(2))=-8pi p+Lambda.]:}\begin{align*} {\left[\frac{\ddot{a}}{a}+2\left(\frac{\dot{a}}{a}\right)^{2}+2 \frac{k}{a}\right]-3\left[\frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}\right] } & =8 \pi p-\Lambda \\ 2 \frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}} & =-8 \pi p+\Lambda . \tag{17.5} \end{align*}[a¨a+2(a˙a)2+2ka]3[a¨a+(a˙a)2+ka2]=8πpΛ(17.5)2a¨a+(a˙a)2+ka2=8πp+Λ.
We are left with two simple equations for the evolution of the Robertson-Walker spaces.
The two equations of motion from the last example for a ( t ) a ( t ) a(t)a(t)a(t) in terms of energy density, pressure, and constant curvature are repeated below, with their usual names.
Friedmann equation 1: the initial value equation
(17.6) ( a ˙ a ) 2 + k a 2 = 8 π 3 ρ + Λ 3 (17.6) a ˙ a 2 + k a 2 = 8 π 3 ρ + Λ 3 {:(17.6)(((a^(˙)))/(a))^(2)+(k)/(a^(2))=(8pi)/(3)rho+(Lambda)/(3):}\begin{equation*} \left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=\frac{8 \pi}{3} \rho+\frac{\Lambda}{3} \tag{17.6} \end{equation*}(17.6)(a˙a)2+ka2=8π3ρ+Λ3
Friedmann equation 2: the dynamic equation
(17.7) 2 a ¨ a + ( a ˙ a ) 2 + k a 2 = 8 π p + Λ (17.7) 2 a ¨ a + a ˙ a 2 + k a 2 = 8 π p + Λ {:(17.7)2((a^(¨)))/(a)+(((a^(˙)))/(a))^(2)+(k)/(a^(2))=-8pi p+Lambda:}\begin{equation*} 2 \frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=-8 \pi p+\Lambda \tag{17.7} \end{equation*}(17.7)2a¨a+(a˙a)2+ka2=8πp+Λ
So far the relationship between ρ ρ rho\rhoρ and p p ppp has not been given. These quantities are linked by an equation of state which is determined by the thermodynamics of the matter comprising the perfect fluid that fills the model universe. Since the scale of space itself depends on the radiuslike expansion factor a ( t ) a ( t ) a(t)a(t)a(t), this will also come into play in determining the volume in which our matter is confined (and therefore also its density). This is our next topic.

17.2 Enter thermodynamics

In order to make further progress, we need to consider how the effective radius of the universe a a aaa is linked to the thermodynamic variables ρ ρ rho\rhoρ and p p ppp. General relativity is not the only place where geometry and mass-energy are linked: they are also tied together in the thermodynamic laws of physics. In the next example, we consider the first law of thermodynamics.

Example 17.3

The first law of thermodynamics says
(17.8) d U = T d S p d V , (17.8) d U = T d S p d V , {:(17.8)dU=TdS-pdV",":}\begin{equation*} \mathrm{d} U=T \mathrm{~d} S-p \mathrm{~d} V, \tag{17.8} \end{equation*}(17.8)dU=T dSp dV,
linking the internal energy U U UUU, temperature T T TTT and entropy S S SSS in a volume V V VVV. We shall often assume an adiabatic evolution of the universe, 4 4 ^(4){ }^{4}4 so we can set d S = 0 d S = 0 dS=0\mathrm{d} S=0dS=0, Taking the volume V = b a ( t ) 3 V = b a ( t ) 3 V=ba(t)^(3)V=b a(t)^{3}V=ba(t)3, where b b bbb is a dimensionless constant, we then have U = ρ V = b ρ a ( t ) 3 U = ρ V = b ρ a ( t ) 3 U=rho V=b rho a(t)^(3)U=\rho V=b \rho a(t)^{3}U=ρV=bρa(t)3. Taking a time derivative of the first law of thermodynamics, we find (cancelling factors of b b bbb )
d U d t = p d V d t d d t ( ρ a 3 ) = p d d t a 3 (17.9) d ρ d t a 3 + 3 ρ a 2 d a d t = 3 p a 2 d a d t d U d t = p d V d t d d t ρ a 3 = p d d t a 3 (17.9) d ρ d t a 3 + 3 ρ a 2 d a d t = 3 p a 2 d a d t {:[(dU)/((d)t)=-p((d)V)/((d)t)],[((d))/((d)t)(rhoa^(3))=-p((d))/((d)t)a^(3)],[(17.9)((d)rho)/((d)t)*a^(3)+3rhoa^(2)((d)a)/((d)t)=-3pa^(2)((d)a)/((d)t)]:}\begin{align*} \frac{\mathrm{d} U}{\mathrm{~d} t} & =-p \frac{\mathrm{~d} V}{\mathrm{~d} t} \\ \frac{\mathrm{~d}}{\mathrm{~d} t}\left(\rho a^{3}\right) & =-p \frac{\mathrm{~d}}{\mathrm{~d} t} a^{3} \\ \frac{\mathrm{~d} \rho}{\mathrm{~d} t} \cdot a^{3}+3 \rho a^{2} \frac{\mathrm{~d} a}{\mathrm{~d} t} & =-3 p a^{2} \frac{\mathrm{~d} a}{\mathrm{~d} t} \tag{17.9} \end{align*}dU dt=p dV dt d dt(ρa3)=p d dta3(17.9) dρ dta3+3ρa2 da dt=3pa2 da dt
or, on rearranging,
(17.10) 1 ρ + p d ρ d t = 3 a d a d t (17.10) 1 ρ + p d ρ d t = 3 a d a d t {:(17.10)(1)/(rho+p)((d)rho)/((d)t)=-(3)/(a)((d)a)/((d)t):}\begin{equation*} \frac{1}{\rho+p} \frac{\mathrm{~d} \rho}{\mathrm{~d} t}=-\frac{3}{a} \frac{\mathrm{~d} a}{\mathrm{~d} t} \tag{17.10} \end{equation*}(17.10)1ρ+p dρ dt=3a da dt
Using the result of the previous example, we have an alternative statement of the first law of thermodynamics, in a form directly applicable to the Friedmann cosmologies.
The first law of thermodynamics can be written as
(17.11) d ρ ρ + p = 3 d a a (17.11) d ρ ρ + p = 3 d a a {:(17.11)(drho)/(rho+p)=-3((d)a)/(a):}\begin{equation*} \frac{\mathrm{d} \rho}{\rho+p}=-3 \frac{\mathrm{~d} a}{a} \tag{17.11} \end{equation*}(17.11)dρρ+p=3 daa
Armed with this, we can show that the two Friedmann equations actually represent the same information.
\curvearrowright Thermodynamics is explored in more detail in Chapplored 39.
4 4 ^(4){ }^{4}4 This amounts to an assumption that the elements of the cosmic fluid remain in thermal equilibrium and there are no phase transitions or shockwaves. In the case of a shockwave, for example, the kinetic energy of the fluid elements can be converted to heat, leading to an increase in entropy.
Example 17.4
Start with a slight rearrangement of the first Friedmann equation
(17.12) a ˙ 2 + k = ( Λ 3 + 8 π ρ 3 ) a 2 (17.12) a ˙ 2 + k = Λ 3 + 8 π ρ 3 a 2 {:(17.12)a^(˙)^(2)+k=((Lambda)/(3)+(8pi rho)/(3))a^(2):}\begin{equation*} \dot{a}^{2}+k=\left(\frac{\Lambda}{3}+\frac{8 \pi \rho}{3}\right) a^{2} \tag{17.12} \end{equation*}(17.12)a˙2+k=(Λ3+8πρ3)a2
and differentiate to find
(17.13) 2 a ˙ a ¨ = 2 a a ˙ ( Λ 3 + 8 π ρ 3 ) + a 2 8 π ρ ˙ 3 . (17.13) 2 a ˙ a ¨ = 2 a a ˙ Λ 3 + 8 π ρ 3 + a 2 8 π ρ ˙ 3 . {:(17.13)2a^(˙)a^(¨)=2aa^(˙)((Lambda)/(3)+(8pi rho)/(3))+a^(2)(8pi(rho^(˙)))/(3).:}\begin{equation*} 2 \dot{a} \ddot{a}=2 a \dot{a}\left(\frac{\Lambda}{3}+\frac{8 \pi \rho}{3}\right)+a^{2} \frac{8 \pi \dot{\rho}}{3} . \tag{17.13} \end{equation*}(17.13)2a˙a¨=2aa˙(Λ3+8πρ3)+a28πρ˙3.
Combine this with the first law of thermodynamics in the form
(17.14) ρ ˙ = 3 ( ρ + p ) a ˙ a , (17.14) ρ ˙ = 3 ( ρ + p ) a ˙ a , {:(17.14)rho^(˙)=-3(rho+p)((a^(˙)))/(a)",":}\begin{equation*} \dot{\rho}=-3(\rho+p) \frac{\dot{a}}{a}, \tag{17.14} \end{equation*}(17.14)ρ˙=3(ρ+p)a˙a,
to give
(17.15) 2 a ¨ a = 2 Λ 3 8 π ρ 3 8 π p (17.15) 2 a ¨ a = 2 Λ 3 8 π ρ 3 8 π p {:(17.15)2((a^(¨)))/(a)=(2Lambda)/(3)-(8pi rho)/(3)-8pi p:}\begin{equation*} 2 \frac{\ddot{a}}{a}=\frac{2 \Lambda}{3}-\frac{8 \pi \rho}{3}-8 \pi p \tag{17.15} \end{equation*}(17.15)2a¨a=2Λ38πρ38πp
Finally, substitute for 8 π ρ / 3 8 π ρ / 3 8pi rho//38 \pi \rho / 38πρ/3 using the first Friedmann equation and we find the second Friedmann equation
(17.16) 2 a ¨ a + ( a ˙ a ) 2 + k a 2 = 8 π p + Λ . (17.16) 2 a ¨ a + a ˙ a 2 + k a 2 = 8 π p + Λ . {:(17.16)2((a^(¨)))/(a)+(((a^(˙)))/(a))^(2)+(k)/(a^(2))=-8pi p+Lambda.:}\begin{equation*} 2 \frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=-8 \pi p+\Lambda . \tag{17.16} \end{equation*}(17.16)2a¨a+(a˙a)2+ka2=8πp+Λ.
Conclusion: The two Friedmann equations are actually telling us the same thing.
The first law of thermodynamics links ρ , p ρ , p rho,p\rho, pρ,p and a a aaa. In order to specify the equation of state, we must decide what sort of matter we wish to model. Two types of matter are discussed in the next section.

17.3 Dust and radiation

All matter contributes to the energy-momentum tensor and hence affects the geometry of the universe. On the length scales over which we are working, the matter is continuous and so is well described as a classical fluid giving rise to an energy-momentum field. We now divide the matter in the fluid that fills the universe into two sorts:
(1) matter with mass: This is also called dust. It is simply massive, continuous matter with p = 0 p = 0 p=0p=0p=0. The mass-energy density of dust is written as ρ d ρ d rho_(d)\rho_{\mathrm{d}}ρd.
(2) massless matter: This is also called radiation, and is assumed to be a gas of photons with uniform energy density ρ r ρ r rho_(r)\rho_{\mathrm{r}}ρr and non-zero pressure.
The total energy density of the universe is a sum ρ = ρ d + ρ r ρ = ρ d + ρ r rho=rho_(d)+rho_(r)\rho=\rho_{\mathrm{d}}+\rho_{\mathrm{r}}ρ=ρd+ρr.
Considering dust first, setting p = 0 p = 0 p=0p=0p=0, the first law of thermodynamics (eqn 17.11) implies the equation of state for dust is
(17.17) ρ d ( t ) a ( t ) 3 (17.17) ρ d ( t ) a ( t ) 3 {:(17.17)rho_(d)(t)prop a(t)^(-3):}\begin{equation*} \rho_{\mathrm{d}}(t) \propto a(t)^{-3} \tag{17.17} \end{equation*}(17.17)ρd(t)a(t)3
or, if we make measurements relative to some state at a time t = t 0 , 5 t = t 0 , 5 t=t_(0),^(5)t=t_{0},{ }^{5}t=t0,5 we have
(17.18) ρ d ( t ) ρ d ( 0 ) = [ a ( t 0 ) a ( t ) ] 3 (17.18) ρ d ( t ) ρ d ( 0 ) = a t 0 a ( t ) 3 {:(17.18)(rho_(d)(t))/(rho_(d)^((0)))=[(a(t_(0)))/(a(t))]^(3):}\begin{equation*} \frac{\rho_{\mathrm{d}}(t)}{\rho_{\mathrm{d}}^{(0)}}=\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{3} \tag{17.18} \end{equation*}(17.18)ρd(t)ρd(0)=[a(t0)a(t)]3
where we use the notation ρ d ( 0 ) = ρ d ( t 0 ) ρ d ( 0 ) = ρ d t 0 rho_(d)^((0))=rho_(d)(t_(0))\rho_{\mathrm{d}}^{(0)}=\rho_{\mathrm{d}}\left(t_{0}\right)ρd(0)=ρd(t0). We deduce 6 ρ d a ( t ) 3 6 ρ d a ( t ) 3 ^(6)rho_(d)prop a(t)^(-3){ }^{6} \rho_{\mathrm{d}} \propto a(t)^{-3}6ρda(t)3
Now for radiation. 7 7 ^(7){ }^{7}7 The equation of state for radiation is a fundamental result of thermodynamics and is given by
(17.19) p = ρ 3 . (17.19) p = ρ 3 . {:(17.19)p=(rho)/(3).:}\begin{equation*} p=\frac{\rho}{3} . \tag{17.19} \end{equation*}(17.19)p=ρ3.
Eqn 17.11 then gives us
(17.20) d ρ ρ = 4 d a a , (17.20) d ρ ρ = 4 d a a , {:(17.20)(drho)/(rho)=-4((d)a)/(a)",":}\begin{equation*} \frac{\mathrm{d} \rho}{\rho}=-4 \frac{\mathrm{~d} a}{a}, \tag{17.20} \end{equation*}(17.20)dρρ=4 daa,
or, again making measurements relative to conditions at t 0 t 0 t_(0)t_{0}t0,
(17.21) ρ r ( t ) ρ r ( 0 ) = [ a ( t 0 ) a ( t ) ] 4 (17.21) ρ r ( t ) ρ r ( 0 ) = a t 0 a ( t ) 4 {:(17.21)(rho_(r)(t))/(rho_(r)^((0)))=[(a(t_(0)))/(a(t))]^(4):}\begin{equation*} \frac{\rho_{\mathrm{r}}(t)}{\rho_{\mathrm{r}}^{(0)}}=\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{4} \tag{17.21} \end{equation*}(17.21)ρr(t)ρr(0)=[a(t0)a(t)]4
where ρ r ( 0 ) = ρ r ( t 0 ) ρ r ( 0 ) = ρ r t 0 rho_(r)^((0))=rho_(r)(t_(0))\rho_{\mathrm{r}}^{(0)}=\rho_{\mathrm{r}}\left(t_{0}\right)ρr(0)=ρr(t0). We deduce 8 ρ r a ( t ) 4 8 ρ r a ( t ) 4 ^(8)rho_(r)prop a(t)^(-4){ }^{8} \rho_{\mathrm{r}} \propto a(t)^{-4}8ρra(t)4. We conclude that in both cases, dust and radiation, an increasing a ( t ) a ( t ) a(t)a(t)a(t) leads to a power-law drop in the energy density of the universe. In other words, the universe becomes emptier as a function of time, but the energy density due to radiation drops off fastest.
Using the thermodynamic information, we can then rewrite the Friedmann equations for a universe filled with a mixture of dust, radiation, and vacuum energy. They become
(17.22) [ a ˙ ( t ) a ( t ) ] 2 + k a ( t ) 2 = 8 π 3 { ρ d ( 0 ) [ a ( t 0 ) a ( t ) ] 3 + ρ r ( 0 ) [ a ( t 0 ) a ( t ) ] 4 } + Λ 3 (17.23) a ¨ a = 4 π 3 { ρ d ( 0 ) [ a ( t 0 ) a ( t ) ] 3 + ρ r ( 0 ) [ a ( t 0 ) a ( t ) ] 4 } + Λ 3 (17.22) a ˙ ( t ) a ( t ) 2 + k a ( t ) 2 = 8 π 3 ρ d ( 0 ) a t 0 a ( t ) 3 + ρ r ( 0 ) a t 0 a ( t ) 4 + Λ 3 (17.23) a ¨ a = 4 π 3 ρ d ( 0 ) a t 0 a ( t ) 3 + ρ r ( 0 ) a t 0 a ( t ) 4 + Λ 3 {:[(17.22)[((a^(˙))(t))/(a(t))]^(2)+(k)/(a(t)^(2))=(8pi)/(3){rho_(d)^((0))[(a(t_(0)))/(a(t))]^(3)+rho_(r)^((0))[(a(t_(0)))/(a(t))]^(4)}+(Lambda)/(3)],[(17.23)((a^(¨)))/(a)=-(4pi)/(3){rho_(d)^((0))[(a(t_(0)))/(a(t))]^(3)+rho_(r)^((0))[(a(t_(0)))/(a(t))]^(4)}+(Lambda)/(3)]:}\begin{gather*} {\left[\frac{\dot{a}(t)}{a(t)}\right]^{2}+\frac{k}{a(t)^{2}}=\frac{8 \pi}{3}\left\{\rho_{\mathrm{d}}^{(0)}\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{3}+\rho_{\mathrm{r}}^{(0)}\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{4}\right\}+\frac{\Lambda}{3}} \tag{17.22}\\ \frac{\ddot{a}}{a}=-\frac{4 \pi}{3}\left\{\rho_{\mathrm{d}}^{(0)}\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{3}+\rho_{\mathrm{r}}^{(0)}\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{4}\right\}+\frac{\Lambda}{3} \tag{17.23} \end{gather*}(17.22)[a˙(t)a(t)]2+ka(t)2=8π3{ρd(0)[a(t0)a(t)]3+ρr(0)[a(t0)a(t)]4}+Λ3(17.23)a¨a=4π3{ρd(0)[a(t0)a(t)]3+ρr(0)[a(t0)a(t)]4}+Λ3
These equations tell us that if we have values for the initial densities ρ ( 0 ) ρ ( 0 ) rho^((0))\rho^{(0)}ρ(0) at some time t 0 t 0 t_(0)t_{0}t0 when we know the value of the radius a = a ( t 0 ) a = a t 0 a=a(t_(0))a=a\left(t_{0}\right)a=a(t0), we can find the behaviour of the effective radius function a ( t ) a ( t ) a(t)a(t)a(t) at all times.
With the Friedmann equations in hand we can continue our survey of possible universes. The Friedmann equations now allow us to fill each of the Robertson-Walker spaces with different sorts of matter (that is, radiation, dust, and vacuum energy) and examine the consequences.

Example 17.5

It is sometimes useful to express the energy density as a fraction of a critical density ρ c ρ c rho_(c)\rho_{c}ρc, defined by 9 9 ^(9){ }^{9}9
(17.25) ρ c = 3 8 π ( a ˙ a ) 2 (17.25) ρ c = 3 8 π ( a ˙ a ) 2 {:(17.25)rho_(c)=(3)/(8pi)(((a^(˙)))/(a))^(2):}\begin{equation*} \rho_{\mathrm{c}}=\frac{3}{8 \pi}\binom{\dot{a}}{a}^{2} \tag{17.25} \end{equation*}(17.25)ρc=38π(a˙a)2
Next, we define a density of vacuum energy ρ Λ = Λ / 8 π ρ Λ = Λ / 8 π rho_(Lambda)=Lambda//8pi\rho_{\Lambda}=\Lambda / 8 \piρΛ=Λ/8π and obtain
(17.27) 1 + k a 2 H 2 = ρ d ρ c + ρ r ρ c + ρ Λ ρ c . (17.27) 1 + k a 2 H 2 = ρ d ρ c + ρ r ρ c + ρ Λ ρ c . {:(17.27)1+(k)/(a^(2)H^(2))=(rho_(d))/(rho_(c))+(rho_(r))/(rho_(c))+(rho_(Lambda))/(rho_(c)).:}\begin{equation*} 1+\frac{k}{a^{2} H^{2}}=\frac{\rho_{d}}{\rho_{\mathrm{c}}}+\frac{\rho_{r}}{\rho_{\mathrm{c}}}+\frac{\rho_{\Lambda}}{\rho_{\mathrm{c}}} . \tag{17.27} \end{equation*}(17.27)1+ka2H2=ρdρc+ρrρc+ρΛρc.
6 6 ^(6){ }^{6}6 We can rationalize this result as follows: in thermal equilibrium, particle number is conserved, and so the density of matter must be inversely proportional to the volume which is proportional to a ( t ) 3 a ( t ) 3 a(t)^(3)a(t)^{3}a(t)3. The energy of the matter is just its time-independent mass. Hence, the energy density ρ d a ( t ) 3 ρ d a ( t ) 3 rho_(d)prop a(t)^(-3)\rho_{\mathrm{d}} \propto a(t)^{-3}ρda(t)3. 7 7 ^(7){ }^{7}7 See also Exercise 17.2.
8 8 ^(8){ }^{8}8 We can rationalize this result as follows: in thermal equilibrium, particle number is conserved, and so the density of photons must be inversely proportional to the volume which is proportional to a ( t ) 3 a ( t ) 3 a(t)^(3)a(t)^{3}a(t)3. However, the energy of each photon is inversely proportional to a ( t ) a ( t ) a(t)a(t)a(t) because of the effect of redshift. Hence, the energy density ρ r a ( t ) 4 ρ r a ( t ) 4 rho_(r)prop a(t)^(-4)\rho_{\mathrm{r}} \propto a(t)^{-4}ρra(t)4.
9 9 ^(9){ }^{9}9 The name is justified below. Note that, equivalently, we could write
(17.24) ρ c = 3 8 π H ( t ) 2 . (17.24) ρ c = 3 8 π H ( t ) 2 . {:(17.24)rho_(c)=(3)/(8pi)H(t)^(2).:}\begin{equation*} \rho_{\mathrm{c}}=\frac{3}{8 \pi} H(t)^{2} . \tag{17.24} \end{equation*}(17.24)ρc=38πH(t)2.
One way forward is to substitute for a 2 a 2 a^(2)a^{2}a2 from eqn 17.25 , giving
(17.28) 1 + k a ˙ 2 = ρ d ρ c + ρ r ρ c + ρ Λ ρ c (17.28) 1 + k a ˙ 2 = ρ d ρ c + ρ r ρ c + ρ Λ ρ c {:(17.28)1+(k)/(a^(˙)^(2))=(rho_(d))/(rho_(c))+(rho_(r))/(rho_(c))+(rho_(Lambda))/(rho_(c)):}\begin{equation*} 1+\frac{k}{\dot{a}^{2}}=\frac{\rho_{d}}{\rho_{\mathrm{c}}}+\frac{\rho_{r}}{\rho_{\mathrm{c}}}+\frac{\rho_{\Lambda}}{\rho_{\mathrm{c}}} \tag{17.28} \end{equation*}(17.28)1+ka˙2=ρdρc+ρrρc+ρΛρc
We conclude that the first Friedmann equation can be rewritten to read
k a ˙ 2 = ρ d ρ c + ρ r ρ c + ρ A ρ c 1 (17.29) = Ω d + Ω r + Ω Λ 1 , k a ˙ 2 = ρ d ρ c + ρ r ρ c + ρ A ρ c 1 (17.29) = Ω d + Ω r + Ω Λ 1 , {:[(k)/(a^(˙)^(2))=(rho_(d))/(rho_(c))+(rho_(r))/(rho_(c))+(rho_(A))/(rho_(c))-1],[(17.29)=Omega_(d)+Omega_(r)+Omega_(Lambda)-1","]:}\begin{align*} \frac{k}{\dot{a}^{2}} & =\frac{\rho_{\mathrm{d}}}{\rho_{\mathrm{c}}}+\frac{\rho_{\mathrm{r}}}{\rho_{\mathrm{c}}}+\frac{\rho_{\mathrm{A}}}{\rho_{\mathrm{c}}}-1 \\ & =\Omega_{\mathrm{d}}+\Omega_{\mathrm{r}}+\Omega_{\Lambda}-1, \tag{17.29} \end{align*}ka˙2=ρdρc+ρrρc+ρAρc1(17.29)=Ωd+Ωr+ΩΛ1,
where in the last step we have written the ratio Ω j = ρ j / ρ c Ω j = ρ j / ρ c Omega_(j)=rho_(j)//rho_(c)\Omega_{j}=\rho_{j} / \rho_{c}Ωj=ρj/ρc. The usefulness of this equation is that it provides a straightforward means of telling us which of the three possible geometries ( k = 0 , ± 1 ) ( k = 0 , ± 1 ) (k=0,+-1)(k=0, \pm 1)(k=0,±1) must result from a universe containing a particular amount of mass-energy. That is, for a total density ratio Ω = Ω d + Ω r + Ω A Ω = Ω d + Ω r + Ω A Omega=Omega_(d)+Omega_(r)+Omega_(A)\Omega=\Omega_{\mathrm{d}}+\Omega_{\mathrm{r}}+\Omega_{\mathrm{A}}Ω=Ωd+Ωr+ΩA we have k / a ˙ 2 = Ω 1 k / a ˙ 2 = Ω 1 k//a^(˙)^(2)=Omega-1k / \dot{a}^{2}=\Omega-1k/a˙2=Ω1, from which (assuming a ˙ > 0 a ˙ > 0 a^(˙) > 0\dot{a}>0a˙>0 ) we can read off the following:
If Ω < 1 Ω < 1 Omega < 1\Omega<1Ω<1 then k = 1 k = 1 k=-1k=-1k=1 and the space is hyperbolic.
  • If Ω = 1 Ω = 1 Omega=1\Omega=1Ω=1 then k = 0 k = 0 k=0k=0k=0 and the space is flat.
  • If Ω > 1 Ω > 1 Omega > 1\Omega>1Ω>1 then k = 1 k = 1 k=1k=1k=1 and the space is spherical.
We can now see why ρ c ρ c rho_(c)\rho_{\mathrm{c}}ρc is called a critical density: it's only if the total density ρ = ρ c ρ = ρ c rho=rho_(c)\rho=\rho_{\mathrm{c}}ρ=ρc that we have the result that k = 0 k = 0 k=0k=0k=0.
Another way of expressing things is to relate everything to measurements taken at t = t 0 t = t 0 t=t_(0)t=t_{0}t=t0, as we did in eqn 17.22 . First we write ρ c ( 0 ) = 8 π H 0 / 3 ρ c ( 0 ) = 8 π H 0 / 3 rho_(c)^((0))=8piH_(0)//3\rho_{\mathrm{c}}^{(0)}=8 \pi H_{0} / 3ρc(0)=8πH0/3 and Ω j ( 0 ) = ρ j ( 0 ) / ρ c ( 0 ) Ω j ( 0 ) = ρ j ( 0 ) / ρ c ( 0 ) Omega_(j)^((0))=rho_(j)^((0))//rho_(c)^((0))\Omega_{j}^{(0)}=\rho_{j}^{(0)} / \rho_{\mathrm{c}}^{(0)}Ωj(0)=ρj(0)/ρc(0). Then define Ω k ( 0 ) = k / a ( t 0 ) 2 H 0 2 Ω k ( 0 ) = k / a t 0 2 H 0 2 Omega_(k)^((0))=-k//a(t_(0))^(2)H_(0)^(2)\Omega_{k}^{(0)}=-k / a\left(t_{0}\right)^{2} H_{0}^{2}Ωk(0)=k/a(t0)2H02, so that we have, from eqn 17.27, that
(17.30) Ω d ( 0 ) + Ω r ( 0 ) + Ω Λ ( 0 ) + Ω k ( 0 ) = 1 (17.30) Ω d ( 0 ) + Ω r ( 0 ) + Ω Λ ( 0 ) + Ω k ( 0 ) = 1 {:(17.30)Omega_(d)^((0))+Omega_(r)^((0))+Omega_(Lambda)^((0))+Omega_(k)^((0))=1:}\begin{equation*} \Omega_{\mathrm{d}}^{(0)}+\Omega_{\mathrm{r}}^{(0)}+\Omega_{\Lambda}^{(0)}+\Omega_{k}^{(0)}=1 \tag{17.30} \end{equation*}(17.30)Ωd(0)+Ωr(0)+ΩΛ(0)+Ωk(0)=1
Returning to eqn 17.22 we then write
(17.31) [ a ˙ ( t ) a ( t ) ] 2 + k a ( t ) 2 = { Ω d ( 0 ) [ a ( t 0 ) a ( t ) ] 3 + Ω r ( 0 ) [ a ( t 0 ) a ( t ) ] 4 + Ω Λ ( 0 ) } (17.31) a ˙ ( t ) a ( t ) 2 + k a ( t ) 2 = Ω d ( 0 ) a t 0 a ( t ) 3 + Ω r ( 0 ) a t 0 a ( t ) 4 + Ω Λ ( 0 ) {:(17.31)[((a^(˙))(t))/(a(t))]^(2)+(k)/(a(t)^(2))={Omega_(d)^((0))[(a(t_(0)))/(a(t))]^(3)+Omega_(r)^((0))[(a(t_(0)))/(a(t))]^(4)+Omega_(Lambda)^((0))}:}\begin{equation*} \left[\frac{\dot{a}(t)}{a(t)}\right]^{2}+\frac{k}{a(t)^{2}}=\left\{\Omega_{\mathrm{d}}^{(0)}\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{3}+\Omega_{\mathrm{r}}^{(0)}\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{4}+\Omega_{\Lambda}^{(0)}\right\} \tag{17.31} \end{equation*}(17.31)[a˙(t)a(t)]2+ka(t)2={Ωd(0)[a(t0)a(t)]3+Ωr(0)[a(t0)a(t)]4+ΩΛ(0)}
Simplifying by writing x = a ( t ) / a ( t 0 ) x = a ( t ) / a t 0 x=a(t)//a(t_(0))x=a(t) / a\left(t_{0}\right)x=a(t)/a(t0) we have
(17.32) 1 x 2 ( d x d t ) 2 = H 0 2 ( Ω k ( 0 ) x 2 + Ω d ( 0 ) x 3 + Ω r ( 0 ) x 4 + Ω Λ ( 0 ) ) (17.32) 1 x 2 d x d t 2 = H 0 2 Ω k ( 0 ) x 2 + Ω d ( 0 ) x 3 + Ω r ( 0 ) x 4 + Ω Λ ( 0 ) {:(17.32)(1)/(x^(2))(((d)x)/((d)t))^(2)=H_(0)^(2)((Omega_(k)^((0)))/(x^(2))+(Omega_(d)^((0)))/(x^(3))+(Omega_(r)^((0)))/(x^(4))+Omega_(Lambda)^((0))):}\begin{equation*} \frac{1}{x^{2}}\left(\frac{\mathrm{~d} x}{\mathrm{~d} t}\right)^{2}=H_{0}^{2}\left(\frac{\Omega_{k}^{(0)}}{x^{2}}+\frac{\Omega_{\mathrm{d}}^{(0)}}{x^{3}}+\frac{\Omega_{\mathrm{r}}^{(0)}}{x^{4}}+\Omega_{\Lambda}^{(0)}\right) \tag{17.32} \end{equation*}(17.32)1x2( dx dt)2=H02(Ωk(0)x2+Ωd(0)x3+Ωr(0)x4+ΩΛ(0))
The usefulness of this expression is that it can be related to light reaching us with some value of redshift z z zzz. If light was emitted at a time t t ttt and observed at the current time, t 0 t 0 t_(0)t_{0}t0, then we have, by eqn 16.48 , that x = a ( t ) / a ( t 0 ) = ( 1 + z ) 1 x = a ( t ) / a t 0 = ( 1 + z ) 1 x=a(t)//a(t_(0))=(1+z)^(-1)x=a(t) / a\left(t_{0}\right)=(1+z)^{-1}x=a(t)/a(t0)=(1+z)1. This means that if we define the zero of time as corresponding to infinite redshift, then the time at which light was emitted that reaches us with redshift z z zzz is given by
(17.33) t ( z ) = 1 H 0 0 ( z + 1 ) 1 d x x ( Ω k ( 0 ) x 2 + Ω d ( 0 ) x 3 + Ω r ( 0 ) x 4 + Ω Λ ( 0 ) ) 1 2 . (17.33) t ( z ) = 1 H 0 0 ( z + 1 ) 1 d x x Ω k ( 0 ) x 2 + Ω d ( 0 ) x 3 + Ω r ( 0 ) x 4 + Ω Λ ( 0 ) 1 2 . {:(17.33)t(z)=(1)/(H_(0))int_(0)^((z+1)^(-1))((d)x)/(x(Omega_(k)^((0))x^(-2)+Omega_(d)^((0))x^(-3)+Omega_(r)^((0))x^(-4)+Omega_(Lambda)^((0)))^((1)/(2))).:}\begin{equation*} t(z)=\frac{1}{H_{0}} \int_{0}^{(z+1)^{-1}} \frac{\mathrm{~d} x}{x\left(\Omega_{\mathrm{k}}^{(0)} x^{-2}+\Omega_{\mathrm{d}}^{(0)} x^{-3}+\Omega_{\mathrm{r}}^{(0)} x^{-4}+\Omega_{\Lambda}^{(0)}\right)^{\frac{1}{2}}} . \tag{17.33} \end{equation*}(17.33)t(z)=1H00(z+1)1 dxx(Ωk(0)x2+Ωd(0)x3+Ωr(0)x4+ΩΛ(0))12.
Notably, if we set z = 0 z = 0 z=0z=0z=0, this equation outputs the present age of the Universe.
We now have equations that link the matter content of the universe and its geometry. This is a central goal of cosmology. We are, therefore, now ready for the payoff in the following chapter: a description of the possible Robertson-Walker universes, their evolution, origin, and fate.

Chapter summary

  • Filling the Robertson-Walker universes with a perfect fluid gives us the Friedmann equations for the evolution of a ( t ) a ( t ) a(t)a(t)a(t) with time. These can be given in terms of the dust, radiation and vacuumenergy content of the universe.
  • The Friedmann equations are given by
(17.34) ( a ˙ a ) 2 + k a 2 = 8 π 3 ρ + Λ 3 , (17.35) 2 a ¨ a + ( a ˙ a ) 2 + k a 2 = 8 π p + Λ . (17.34) a ˙ a 2 + k a 2 = 8 π 3 ρ + Λ 3 , (17.35) 2 a ¨ a + a ˙ a 2 + k a 2 = 8 π p + Λ . {:[(17.34)(((a^(˙)))/(a))^(2)+(k)/(a^(2))=(8pi)/(3)*rho+(Lambda)/(3)","],[(17.35)2((a^(¨)))/(a)+(((a^(˙)))/(a))^(2)+(k)/(a^(2))=-8pi p+Lambda.]:}\begin{gather*} \left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=\frac{8 \pi}{3} \cdot \rho+\frac{\Lambda}{3}, \tag{17.34}\\ 2 \frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=-8 \pi p+\Lambda . \tag{17.35} \end{gather*}(17.34)(a˙a)2+ka2=8π3ρ+Λ3,(17.35)2a¨a+(a˙a)2+ka2=8πp+Λ.
  • The energy density of matter ρ d a ( t ) 3 ρ d a ( t ) 3 rho_(d)prop a(t)^(-3)\rho_{\mathrm{d}} \propto a(t)^{-3}ρda(t)3 and the energy density of radiation ρ r a ( t ) 4 ρ r a ( t ) 4 rho_(r)prop a(t)^(-4)\rho_{\mathrm{r}} \propto a(t)^{-4}ρra(t)4.

Exercises

(17.1) We had in Chapter 10 that T χ ^ χ ^ = p T χ ^ χ ^ = p T_( hat(chi) hat(chi))=pT_{\hat{\chi} \hat{\chi}}=pTχ^χ^=p
(a) Using the transformation from the previous chapter that said d χ 2 = d r 2 / ( 1 k r 2 ) d χ 2 = d r 2 / 1 k r 2 dchi^(2)=dr^(2)//(1-kr^(2))\mathrm{d} \chi^{2}=\mathrm{d} r^{2} /\left(1-k r^{2}\right)dχ2=dr2/(1kr2), compute
Λ χ r = χ / r Λ χ r = χ / r Lambda^(chi)_(r)=del chi//del r\Lambda^{\chi}{ }_{r}=\partial \chi / \partial rΛχr=χ/r.
(b) Verify that
( e χ ) x ¯ ( e r ^ ) r = ( 1 k r 2 ) 1 2 e χ x ¯ e r ^ r = 1 k r 2 1 2 (e_(chi))^( bar(x))(e_( hat(r)))^(r)=(1-kr^(2))^((1)/(2))\left(\boldsymbol{e}_{\chi}\right)^{\bar{x}}\left(\boldsymbol{e}_{\hat{r}}\right)^{r}=\left(1-k r^{2}\right)^{\frac{1}{2}}(eχ)x¯(er^)r=(1kr2)12
(c) Use the previous results to show that T r ^ r ^ = T r ^ r ^ = T_( hat(r) hat(r))=T_{\hat{r} \hat{r}}=Tr^r^= T χ ^ χ ^ = p T χ ^ χ ^ = p T_( hat(chi) hat(chi))=pT_{\hat{\chi} \hat{\chi}}=pTχ^χ^=p.
(17.2) We can check the thermodynamic results for photons using statistical mechanics. The single particle partition function for a gas of ultra-relativistic indistinguishable particles (i.e. photons) in a 3-
volume V V VVV is
(17.36) Z = V π 2 ( k B T ) 3 (17.36) Z = V π 2 k B T 3 {:(17.36)Z=(V)/(pi^(2))((k_(B)T)/(ℏ))^(3):}\begin{equation*} Z=\frac{V}{\pi^{2}}\left(\frac{k_{\mathrm{B}} T}{\hbar}\right)^{3} \tag{17.36} \end{equation*}(17.36)Z=Vπ2(kBT)3
where T T TTT is the temperature.
(a) Compute the internal energy of the gas.
(b) Compute its pressure.
(c) Verify that p = ρ / 3 p = ρ / 3 p=rho//3p=\rho / 3p=ρ/3 as claimed in the chapter.
(17.3) Consider a flat universe filled only with matter with an equation of state ρ a ( t ) 2 ρ a ( t ) 2 rho prop a(t)^(-2)\rho \propto a(t)^{-2}ρa(t)2. How does a ( t ) a ( t ) a(t)a(t)a(t) evolve?
(17.4) For a flat, dust-dominated model universe, calculate the age of the universe when an observed photon was emitted from a galaxy with redshift z z zzz.

18

18.1 Spatially flat universes 188 18.2 Curved universes with Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0 191
18.3 Einstein, Lemaître and Eddington
Friedmann equation 1:
( a ˙ a ) 2 + k a 2 = 8 π 3 ρ + Λ 3 a ˙ a 2 + k a 2 = 8 π 3 ρ + Λ 3 (((a^(˙)))/(a))^(2)+(k)/(a^(2))=(8pi)/(3)*rho+(Lambda)/(3)\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=\frac{8 \pi}{3} \cdot \rho+\frac{\Lambda}{3}(a˙a)2+ka2=8π3ρ+Λ3
Friedmann equation 2:
2 a ¨ a + ( a ˙ a ) 2 + k a 2 = 8 π p + Λ . 2 a ¨ a + a ˙ a 2 + k a 2 = 8 π p + Λ . 2((a^(¨)))/(a)+(((a^(˙)))/(a))^(2)+(k)/(a^(2))=-8pi p+Lambda.2 \frac{\ddot{a}}{a}+\left(\frac{\dot{a}}{a}\right)^{2}+\frac{k}{a^{2}}=-8 \pi p+\Lambda .2a¨a+(a˙a)2+ka2=8πp+Λ.
Recall that we also define a Hubble parameter
H ( t ) = a ˙ ( t ) a ( t ) . H ( t ) = a ˙ ( t ) a ( t ) . H(t)=((a^(˙))(t))/(a(t)).H(t)=\frac{\dot{a}(t)}{a(t)} .H(t)=a˙(t)a(t).
This is a useful quantity as it tells us the rate of change of proper distance l ( t ) l ( t ) l(t)l(t)l(t) via l ( t ) = H ( t ) l ( t ) l ( t ) = H ( t ) l ( t ) l(t)=H(t)l(t)l(t)=H(t) l(t)l(t)=H(t)l(t).

Universes of the past and future

We will now discuss in a little more detail the struggle for existence
Charles Darwin (1809-1882)
After determining the geometry of the Robertson-Walker spaces and combining them with matter to derive the Friedmann equations, we now continue our survey of possible universes. Each of the models we discuss in this chapter makes use of the metric field described by the Robertson-Walker line element
(18.1) d s 2 = d t 2 + a ( t ) 2 [ d r 2 1 k r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] . (18.1) d s 2 = d t 2 + a ( t ) 2 d r 2 1 k r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 . {:(18.1)ds^(2)=-dt^(2)+a(t)^(2)[((d)r^(2))/(1-kr^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))].:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\frac{\mathrm{~d} r^{2}}{1-k r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] . \tag{18.1} \end{equation*}(18.1)ds2=dt2+a(t)2[ dr21kr2+r2( dθ2+sin2θ dϕ2)].
We choose to fill the universes with perfect fluid and vacuum energy, so can use the Friedmann equations to compute the behaviour of the expansion factor a ( t ) a ( t ) a(t)a(t)a(t), which can be thought of as the effective radius of the universe. The universes can be broadly classified by the value of the curvature constant k k kkk that they take. [Recall that the two universes (numbered 0 and 1) that we have already examined had k = 0 k = 0 k=0k=0k=0 and so were spatially flat.] We start this chapter by exploring some more spatially flat universes before going on to look at some k = 1 k = 1 k=1k=1k=1 and k = 1 k = 1 k=-1k=-1k=1 cases. Throughout we assume that we know the value of the expansion factor a ( t = t 0 ) a t = t 0 a(t=t_(0))a\left(t=t_{0}\right)a(t=t0) at some time t 0 t 0 t_(0)t_{0}t0.

18.1 Spatially flat universes

We start by filling spatially flat ( k = 0 k = 0 k=0k=0k=0 ) universes with different sorts of matter. Recall that Universe 0 0 0\mathbf{0}0 is empty and flat. As a result, it is identical to static, Minkowski spacetime. Universe 1 (the de Sitter model) is flat by construction, and contains only vacuum energy Λ Λ Lambda\LambdaΛ. The result is that this latter universe expands with a ( t ) = a ( t 0 ) e H ( t t 0 ) a ( t ) = a t 0 e H t t 0 a(t)=a(t_(0))e^(H(t-t_(0)))a(t)=a\left(t_{0}\right) \mathrm{e}^{H\left(t-t_{0}\right)}a(t)=a(t0)eH(tt0), where the Hubble parameter H a ˙ / a = Λ / 3 H a ˙ / a = Λ / 3 H-=a^(˙)//a=sqrtLambda//3H \equiv \dot{a} / a=\sqrt{\Lambda / 3}Ha˙/a=Λ/3 is a constant. This universe is not badly behaved at t = 0 t = 0 t=0t=0t=0, where we expect a non-zero radius a ( t = 0 ) a ( t = 0 ) a(t=0)a(t=0)a(t=0) and finite mass-energy density ρ ( t = 0 ) ρ ( t = 0 ) rho(t=0)\rho(t=0)ρ(t=0). Universe 1 does not, therefore, have any singularity at t = 0 t = 0 t=0t=0t=0 and so has no Big Bang.
Now let's consider models without any vacuum energy, but where we allow cosmological fluid with a time-dependent energy density ρ ( t ) ρ ( t ) rho(t)\rho(t)ρ(t) to fill a flat universe. We then have (using the first Friedmann equation,
with k = Λ = 0 k = Λ = 0 k=Lambda=0k=\Lambda=0k=Λ=0 ) that
(18.2) [ a ˙ ( t ) a ( t ) ] 2 = 8 π ρ ( t ) 3 = H ( t ) 2 (18.2) a ˙ ( t ) a ( t ) 2 = 8 π ρ ( t ) 3 = H ( t ) 2 {:(18.2)[((a^(˙))(t))/(a(t))]^(2)=(8pi rho(t))/(3)=H(t)^(2):}\begin{equation*} \left[\frac{\dot{a}(t)}{a(t)}\right]^{2}=\frac{8 \pi \rho(t)}{3}=H(t)^{2} \tag{18.2} \end{equation*}(18.2)[a˙(t)a(t)]2=8πρ(t)3=H(t)2
where now the Hubble parameter is time dependent and is given by H ( t ) = [ 8 π ρ ( t ) / 3 ] 1 2 H ( t ) = [ 8 π ρ ( t ) / 3 ] 1 2 H(t)=[8pi rho(t)//3]^((1)/(2))H(t)=[8 \pi \rho(t) / 3]^{\frac{1}{2}}H(t)=[8πρ(t)/3]12. We saw in the last chapter that it is thermodynamic considerations that cause the energy density ρ ρ rho\rhoρ to be time dependent and that ρ ( t ) ρ ( t ) rho(t)\rho(t)ρ(t) is itself a function of a ( t ) a ( t ) a(t)a(t)a(t). This is to say that, as the universe evolves, the energy density ρ ρ rho\rhoρ changes as the expansion factor a ( t ) a ( t ) a(t)a(t)a(t) changes. Let's use these features to create two new, spatially flat universes. We shall fill them, not with vacuum energy density, but with either radiation (Universe 2) or dust (Universe 3).
Universe 2 is a flat, radiation-filled model. That is, we fill the universe with radiation, which has an energy density
(18.3) ρ r ( t ) = [ a ( t 0 ) a ( t ) ] 4 ρ r ( 0 ) (18.3) ρ r ( t ) = a t 0 a ( t ) 4 ρ r ( 0 ) {:(18.3)rho_(r)(t)=[(a(t_(0)))/(a(t))]^(4)rho_(r)^((0)):}\begin{equation*} \rho_{\mathrm{r}}(t)=\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{4} \rho_{\mathrm{r}}^{(0)} \tag{18.3} \end{equation*}(18.3)ρr(t)=[a(t0)a(t)]4ρr(0)
where measurements are made relative to the state ρ r ( 0 ) ρ r ( 0 ) rho_(r)^((0))\rho_{\mathrm{r}}^{(0)}ρr(0) that occurs at time t 0 t 0 t_(0)t_{0}t0. The Friedmann equation becomes
(18.4) [ a ˙ ( t ) a ( t ) ] 2 = 8 π ρ r ( 0 ) 3 [ a ( t 0 ) a ( t ) ] 4 (18.4) a ˙ ( t ) a ( t ) 2 = 8 π ρ r ( 0 ) 3 a t 0 a ( t ) 4 {:(18.4)[((a^(˙))(t))/(a(t))]^(2)=(8pirho_(r)^((0)))/(3)*[(a(t_(0)))/(a(t))]^(4):}\begin{equation*} \left[\frac{\dot{a}(t)}{a(t)}\right]^{2}=\frac{8 \pi \rho_{\mathrm{r}}^{(0)}}{3} \cdot\left[\frac{a\left(t_{0}\right)}{a(t)}\right]^{4} \tag{18.4} \end{equation*}(18.4)[a˙(t)a(t)]2=8πρr(0)3[a(t0)a(t)]4
For ease of writing, let's also write a ( t 0 ) = a 0 a t 0 = a 0 a(t_(0))=a_(0)a\left(t_{0}\right)=a_{0}a(t0)=a0, and then we have
(18.5) a ˙ ( t ) = H 0 a 0 2 a ( t ) (18.5) a ˙ ( t ) = H 0 a 0 2 a ( t ) {:(18.5)a^(˙)(t)=(H_(0)a_(0)^(2))/(a(t)):}\begin{equation*} \dot{a}(t)=\frac{H_{0} a_{0}^{2}}{a(t)} \tag{18.5} \end{equation*}(18.5)a˙(t)=H0a02a(t)
where H 0 = [ 8 π ρ r ( 0 ) / 3 ] 1 2 H 0 = 8 π ρ r ( 0 ) / 3 1 2 H_(0)=[8pirho_(r)^((0))//3]^((1)/(2))H_{0}=\left[8 \pi \rho_{\mathrm{r}}^{(0)} / 3\right]^{\frac{1}{2}}H0=[8πρr(0)/3]12. This equation is solved by an expansion factor
(18.6) a ( t ) = a 0 2 H 0 t (18.6) a ( t ) = a 0 2 H 0 t {:(18.6)a(t)=a_(0)sqrt(2H_(0)t):}\begin{equation*} a(t)=a_{0} \sqrt{2 H_{0} t} \tag{18.6} \end{equation*}(18.6)a(t)=a02H0t
This spatially flat, expanding universe grows without bound as illustrated in Fig. 18.1. Notice also that at t = 0 t = 0 t=0t=0t=0 we have a ( t = 0 ) = 0 a ( t = 0 ) = 0 a(t=0)=0a(t=0)=0a(t=0)=0. That is, the universe begins at zero spatial size, implying an initially infinite density of the radiation field. This is the Big Bang: an initial singularity in the model of the universe. We confirm that the density is indeed infinite in the next example.

Example 18.1

With the solution in eqn 18.5, we can evaluate the Hubble parameter H ( t ) = a ˙ ( t ) a ( t ) H ( t ) = a ˙ ( t ) a ( t ) H(t)=((a^(˙))(t))/(a(t))H(t)=\frac{\dot{a}(t)}{a(t)}H(t)=a˙(t)a(t). Plugging in, we find
(18.7) H ( t ) = 1 2 a 0 ( 2 H 0 / t ) 1 / 2 a 0 ( 2 H 0 t ) 1 / 2 = 1 2 t (18.7) H ( t ) = 1 2 a 0 2 H 0 / t 1 / 2 a 0 2 H 0 t 1 / 2 = 1 2 t {:(18.7)H(t)=((1)/(2)a_(0)(2H_(0)//t)^(1//2))/(a_(0)(2H_(0)t)^(1//2))=(1)/(2t):}\begin{equation*} H(t)=\frac{\frac{1}{2} a_{0}\left(2 H_{0} / t\right)^{1 / 2}}{a_{0}\left(2 H_{0} t\right)^{1 / 2}}=\frac{1}{2 t} \tag{18.7} \end{equation*}(18.7)H(t)=12a0(2H0/t)1/2a0(2H0t)1/2=12t
So we have that H 0 = 1 / 2 t 0 H 0 = 1 / 2 t 0 H_(0)=1//2t_(0)H_{0}=1 / 2 t_{0}H0=1/2t0, and that the Hubble parameter (i) is badly behaved at t = 0 t = 0 t=0t=0t=0, and (ii) decreases as the universe expands. This is also the case for ρ ( t ) ρ ( t ) rho(t)\rho(t)ρ(t) [since H ( t ) ρ ( t ) 1 2 H ( t ) ρ ( t ) 1 2 H(t)prop rho(t)^((1)/(2))H(t) \propto \rho(t)^{\frac{1}{2}}H(t)ρ(t)12 ], and so we deduce that, for Universe 2, the energy density starts infinite and decreases with time to zero.
Fig. 18.1 The evolution of Universes 2 and 3, which have k = 0 k = 0 k=0k=0k=0 and Λ = Λ = Lambda=\Lambda=Λ= 0 . These spatially flat universes expand from an initial Big-Bang singularity without bound. The flat planes represent flat three-dimensional space.
Universe 3 is also known as the Einstein-de Sitter model. This spatially flat ( k = 0 ) ( k = 0 ) (k=0)(k=0)(k=0) universe is filled only with dust, with energy density
(18.8) ρ d ( t ) = [ a 0 a ( t ) ] 3 ρ d ( 0 ) (18.8) ρ d ( t ) = a 0 a ( t ) 3 ρ d ( 0 ) {:(18.8)rho_(d)(t)=[(a_(0))/(a(t))]^(3)rho_(d)^((0)):}\begin{equation*} \rho_{\mathrm{d}}(t)=\left[\frac{a_{0}}{a(t)}\right]^{3} \rho_{\mathrm{d}}^{(0)} \tag{18.8} \end{equation*}(18.8)ρd(t)=[a0a(t)]3ρd(0)
With this equation of state, we have a Friedmann equation
(18.9) a ˙ = H 0 a 0 3 2 a 1 2 (18.9) a ˙ = H 0 a 0 3 2 a 1 2 {:(18.9)a^(˙)=(H_(0)a_(0)^((3)/(2)))/(a^((1)/(2))):}\begin{equation*} \dot{a}=\frac{H_{0} a_{0}^{\frac{3}{2}}}{a^{\frac{1}{2}}} \tag{18.9} \end{equation*}(18.9)a˙=H0a032a12
implying that
(18.10) a ( t ) = a 0 [ 3 2 H 0 t ] 2 3 (18.10) a ( t ) = a 0 3 2 H 0 t 2 3 {:(18.10)a(t)=a_(0)[(3)/(2)H_(0)t]^((2)/(3)):}\begin{equation*} a(t)=a_{0}\left[\frac{3}{2} H_{0} t\right]^{\frac{2}{3}} \tag{18.10} \end{equation*}(18.10)a(t)=a0[32H0t]23
From this we find that H ( t ) a ˙ / a = 2 / ( 3 t ) H ( t ) a ˙ / a = 2 / ( 3 t ) H(t)-=a^(˙)//a=2//(3t)H(t) \equiv \dot{a} / a=2 /(3 t)H(t)a˙/a=2/(3t). Again, the universe starts from a Big-Bang singularity and grows without bound, evolving as shown in Fig. 18.1.
We have now examined three spatially flat models filled with different manifestations of energy. In each case, the universe is found to expand without limits. We summarize our findings in the table below:
de Sitter Radiation Einstein-de Sitter
Composition Energy Radiation Dust
a ( t ) a ( t ) a(t)a(t)a(t) a 0 e H 0 ( t t 0 ) a 0 e H 0 t t 0 a_(0)e^(H_(0)(t-t_(0)))a_{0} \mathrm{e}^{H_{0}\left(t-t_{0}\right)}a0eH0(tt0) a 0 ( 2 H 0 t ) 1 2 a 0 2 H 0 t 1 2 a_(0)(2H_(0)t)^((1)/(2))a_{0}\left(2 H_{0} t\right)^{\frac{1}{2}}a0(2H0t)12 a 0 ( 3 H 0 t / 2 ) 2 3 a 0 3 H 0 t / 2 2 3 a_(0)(3H_(0)t//2)^((2)/(3))a_{0}\left(3 H_{0} t / 2\right)^{\frac{2}{3}}a0(3H0t/2)23
H ( t ) H ( t ) H(t)H(t)H(t) const. 1 / ( 2 t ) 1 / ( 2 t ) 1//(2t)1 /(2 t)1/(2t) 2 / ( 3 t ) 2 / ( 3 t ) 2//(3t)2 /(3 t)2/(3t)
Big Bang No Yes Yes
de Sitter Radiation Einstein-de Sitter Composition Energy Radiation Dust a(t) a_(0)e^(H_(0)(t-t_(0))) a_(0)(2H_(0)t)^((1)/(2)) a_(0)(3H_(0)t//2)^((2)/(3)) H(t) const. 1//(2t) 2//(3t) Big Bang No Yes Yes| | de Sitter | Radiation | Einstein-de Sitter | | :---: | :---: | :---: | :---: | | Composition | Energy | Radiation | Dust | | $a(t)$ | $a_{0} \mathrm{e}^{H_{0}\left(t-t_{0}\right)}$ | $a_{0}\left(2 H_{0} t\right)^{\frac{1}{2}}$ | $a_{0}\left(3 H_{0} t / 2\right)^{\frac{2}{3}}$ | | $H(t)$ | const. | $1 /(2 t)$ | $2 /(3 t)$ | | Big Bang | No | Yes | Yes |
The evolution of a ( t ) a ( t ) a(t)a(t)a(t) for each model is compared in Fig. 18.2. Let's return to Universe 3 [the ( k = 0 ) ( k = 0 ) (k=0)(k=0)(k=0) Einstein-de Sitter model], which contains only dust, and add in some vacuum energy Λ Λ Lambda\LambdaΛ. The first Friedmann equation for this flat, dust-and-vacuum-energy-universe (Universe 4) now takes the form 1 1 ^(1){ }^{1}1
(18.12) a ˙ 2 = C a + Λ 3 a 2 (18.12) a ˙ 2 = C a + Λ 3 a 2 {:(18.12)a^(˙)^(2)=(C)/(a)+(Lambda)/(3)a^(2):}\begin{equation*} \dot{a}^{2}=\frac{C}{a}+\frac{\Lambda}{3} a^{2} \tag{18.12} \end{equation*}(18.12)a˙2=Ca+Λ3a2
where C = 8 π 3 a 0 3 ρ d ( 0 ) C = 8 π 3 a 0 3 ρ d ( 0 ) C=(8pi)/(3)*a_(0)^(3)rho_(d)^((0))C=\frac{8 \pi}{3} \cdot a_{0}^{3} \rho_{\mathrm{d}}^{(0)}C=8π3a03ρd(0), and we assume that the vacuum energy Λ > 0 Λ > 0 Lambda > 0\Lambda>0Λ>0. In the initial stages of the expansion, where the radius a ( t ) a ( t ) a(t)a(t)a(t) of the universe is small, we expect the dust term to be large compared to the vacuum energy term and so the universe has the behaviour described above for Universe 3, with a t 2 / 3 a t 2 / 3 a propt^(2//3)a \propto t^{2 / 3}at2/3 (implying a Big Bang). However, with Λ > Λ > Lambda >\Lambda>Λ> 0 , at some point when a ( t ) a ( t ) a(t)a(t)a(t) is large enough, the vacuum energy must dominate and we have the behaviour of the de Sitter universe (Universe 1) that a e H t a e H t a prope^(Ht)a \propto \mathrm{e}^{H t}aeHt. Actually, the full behaviour of Universe 4 can be calculated exactly, with the result that
(18.13) a ( t ) = ( 3 C 2 Λ ) 1 3 [ cosh 3 Λ t 1 ] 1 3 (18.13) a ( t ) = 3 C 2 Λ 1 3 [ cosh 3 Λ t 1 ] 1 3 {:(18.13)a(t)=((3C)/(2Lambda))^((1)/(3))[cosh sqrt(3Lambda)t-1]^((1)/(3)):}\begin{equation*} a(t)=\left(\frac{3 C}{2 \Lambda}\right)^{\frac{1}{3}}[\cosh \sqrt{3 \Lambda} t-1]^{\frac{1}{3}} \tag{18.13} \end{equation*}(18.13)a(t)=(3C2Λ)13[cosh3Λt1]13
The general behaviour of this universe is therefore that it starts from a singularity and expands without bound.
Finally, we can consider the intriguing possibility (Universe 5) that the universe is filled with vacuum energy with a negative cosmological constant (i.e. Λ < 0 Λ < 0 Lambda < 0\Lambda<0Λ<0 ). Recall that positive Λ Λ Lambda\LambdaΛ vacuum energy exerts a negative pressure, so having Λ < 0 Λ < 0 Lambda < 0\Lambda<0Λ<0 should be expected to provide something closer to positive-pressure mass energy, which causes gravitational attraction, and hence arrests the expansion of the universe if it is large enough. This does indeed happen: the right-hand side of the Friedmann equation vanishes when a ( t ) 3 = 3 C / ( Λ ) a ( t ) 3 = 3 C / ( Λ ) a(t)^(3)=3C//(-Lambda)a(t)^{3}=3 C /(-\Lambda)a(t)3=3C/(Λ) and the expansion must stop. The full behaviour of the model is given by the exact solution
(18.14) a ( t ) = [ 3 C 2 ( Λ ) ] 1 3 [ 1 cos 3 ( Λ ) t ] 1 3 . (18.14) a ( t ) = 3 C 2 ( Λ ) 1 3 [ 1 cos 3 ( Λ ) t ] 1 3 . {:(18.14)a(t)=[(3C)/(2(-Lambda))]^((1)/(3))[1-cos sqrt(3(-Lambda))t]^((1)/(3)).:}\begin{equation*} a(t)=\left[\frac{3 C}{2(-\Lambda)}\right]^{\frac{1}{3}}[1-\cos \sqrt{3(-\Lambda)} t]^{\frac{1}{3}} . \tag{18.14} \end{equation*}(18.14)a(t)=[3C2(Λ)]13[1cos3(Λ)t]13.
This solution tells us that the universe initially expands from a Big-Bang singularity, reaches a maximum radius, and then contracts, ending in the possibility of a final Big-Crunch singularity as shown in Fig. 18.3.

18.2 Curved universes with Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0

So far, all of the universes we have considered are spatially flat, with k = 0 k = 0 k=0k=0k=0. We now turn to some spatially curved cases, and we'll start with examining the behaviour of dust-filled universes with Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0. Equation 18.11, then gives us a Friedmann equation
(18.15) a ˙ 2 + k = C a (18.15) a ˙ 2 + k = C a {:(18.15)a^(˙)^(2)+k=(C)/(a):}\begin{equation*} \dot{a}^{2}+k=\frac{C}{a} \tag{18.15} \end{equation*}(18.15)a˙2+k=Ca
We already have the Einstein-de Sitter model of Universe 3 for the flat ( k = 0 ) ( k = 0 ) (k=0)(k=0)(k=0) case, where the Universe was found to expand without bound from a singularity as shown in Fig. 18.1. What about the cases k = ± 1 k = ± 1 k=+-1k= \pm 1k=±1 ? First, consider the case that the space is spherical ( k = 1 ) ( k = 1 ) (k=1)(k=1)(k=1). This is Universe 6 ( k = 1 , Λ = 0 ) 6 ( k = 1 , Λ = 0 ) 6(k=1,Lambda=0)6(k=1, \Lambda=0)6(k=1,Λ=0), examined in the next example.

Example 18.2

Start with a ˙ 2 = C / a 1 a ˙ 2 = C / a 1 a^(˙)^(2)=C//a-1\dot{a}^{2}=C / a-1a˙2=C/a1 and make a change of variable u 2 = a / C u 2 = a / C u^(2)=a//Cu^{2}=a / Cu2=a/C and the equation of motion becomes
(18.16) u ˙ 2 = 1 4 C 2 u 2 ( 1 u 2 1 ) (18.16) u ˙ 2 = 1 4 C 2 u 2 1 u 2 1 {:(18.16)u^(˙)^(2)=(1)/(4C^(2)u^(2))*((1)/(u^(2))-1):}\begin{equation*} \dot{u}^{2}=\frac{1}{4 C^{2} u^{2}} \cdot\left(\frac{1}{u^{2}}-1\right) \tag{18.16} \end{equation*}(18.16)u˙2=14C2u2(1u21)
We shall integrate this equation, employing so-called Big-Bang initial conditions that a = 0 a = 0 a=0a=0a=0 when t = 0 t = 0 t=0t=0t=0, and hence u ( 0 ) = 0 u ( 0 ) = 0 u(0)=0u(0)=0u(0)=0, and we find
(18.17) 2 0 u ( t ) u 2 d u ( 1 u 2 ) 1 2 = t C (18.17) 2 0 u ( t ) u 2 d u 1 u 2 1 2 = t C {:(18.17)2int_(0)^(u(t))(u^(2)(d)u)/((1-u^(2))^((1)/(2)))=(t)/(C):}\begin{equation*} 2 \int_{0}^{u(t)} \frac{u^{2} \mathrm{~d} u}{\left(1-u^{2}\right)^{\frac{1}{2}}}=\frac{t}{C} \tag{18.17} \end{equation*}(18.17)20u(t)u2 du(1u2)12=tC
The left-hand side can be integrated with the result
(18.18) 2 0 u ( t ) u 2 d u ( 1 u 2 ) 1 2 = sin 1 u ( t ) u ( t ) [ 1 u 2 ( t ) ] 1 2 (18.18) 2 0 u ( t ) u 2 d u 1 u 2 1 2 = sin 1 u ( t ) u ( t ) 1 u 2 ( t ) 1 2 {:(18.18)2int_(0)^(u(t))(u^(2)(d)u)/((1-u^(2))^((1)/(2)))=sin^(-1)u(t)-u(t)[1-u^(2)(t)]^((1)/(2)):}\begin{equation*} 2 \int_{0}^{u(t)} \frac{u^{2} \mathrm{~d} u}{\left(1-u^{2}\right)^{\frac{1}{2}}}=\sin ^{-1} u(t)-u(t)\left[1-u^{2}(t)\right]^{\frac{1}{2}} \tag{18.18} \end{equation*}(18.18)20u(t)u2 du(1u2)12=sin1u(t)u(t)[1u2(t)]12
Fig. 18.3 The evolution of Universe 5 which has k = 0 k = 0 k=0k=0k=0 and negative vacuum energy.
(a) a ( t ) a ( t ) a(t)a(t)a(t)

(c) a ( t ) a ( t ) a(t)a(t)a(t)
Fig. 18.4 The radius function a ( t ) a ( t ) a(t)a(t)a(t) for the Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0, dust-filled universes. (a) Universe 6 with k = 1 k = 1 k=1k=1k=1; (b) Universe 3 (the Einstein-de Sitter model) with k = 0 k = 0 k=0k=0k=0; (c) Universe 7 with k = 1 k = 1 k=-1k=-1k=1.
Fig. 18.5 The evolution of Universe 6 , which has k = 1 k = 1 k=1k=1k=1. The circles represent 3 -spheres.
Fig. 18.6 The evolution of Univig. 7 7 7\mathbf{~ 7} 7, which has k = 1 k = 1 k=-1k=-1k=1 and Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0. The surfaces represent three Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0. The surfaces represent
dimensional hyperbolic space.
We conclude that
(18.19) t = C [ sin 1 ( a C ) 1 2 ( a C ) 1 2 ( 1 a C ) 1 2 ] (18.19) t = C sin 1 a C 1 2 a C 1 2 1 a C 1 2 {:(18.19)t=C[sin^(-1)((a)/(C))^((1)/(2))-((a)/(C))^((1)/(2))(1-(a)/(C))^((1)/(2))]:}\begin{equation*} t=C\left[\sin ^{-1}\left(\frac{a}{C}\right)^{\frac{1}{2}}-\left(\frac{a}{C}\right)^{\frac{1}{2}}\left(1-\frac{a}{C}\right)^{\frac{1}{2}}\right] \tag{18.19} \end{equation*}(18.19)t=C[sin1(aC)12(aC)12(1aC)12]
The behaviour of a ( t ) a ( t ) a(t)a(t)a(t) is shown in Fig. 18.4(a), where we see it expand from zero, reach a maximum and finally collapse to a zero.
The result of the last example is that Universe 6 starts from a Big Bang, expands and then collapses into a big crunch as illustrated in Fig. 18.5, where each of the spatial hypersurfaces represents a 3 -sphere.
For Universe 7, we carry out an analogous computation for a spatially hyperbolic case ( k = 1 ) ( k = 1 ) (k=-1)(k=-1)(k=1). We find that
(18.20) t = C [ ( a C ) 1 2 ( 1 + a C ) 1 2 sinh 1 ( a C ) 1 2 ] (18.20) t = C a C 1 2 1 + a C 1 2 sinh 1 a C 1 2 {:(18.20)t=C[((a)/(C))^((1)/(2))(1+(a)/(C))^((1)/(2))-sinh^(-1)((a)/(C))^((1)/(2))]:}\begin{equation*} t=C\left[\left(\frac{a}{C}\right)^{\frac{1}{2}}\left(1+\frac{a}{C}\right)^{\frac{1}{2}}-\sinh ^{-1}\left(\frac{a}{C}\right)^{\frac{1}{2}}\right] \tag{18.20} \end{equation*}(18.20)t=C[(aC)12(1+aC)12sinh1(aC)12]
This equation, which is graphed in Fig 18.4(c), tells us that Universe 7 grows without bound. [It can be compared with the slower expansion of the Einstein-de Sitter universe shown in Fig. 18.4(b).] Universe 7 is also represented in Fig. 18.6, where each of the spatial hypersurfaces represents a hyperbolic space.
To summarize, the Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0, dust-filled universes we have examined all have an initial expansion from a Big Bang. The flat ( k = 0 ) ( k = 0 ) (k=0)(k=0)(k=0) case (Universe 3) and the hyperbolic k = 1 k = 1 k=-1k=-1k=1 (Universe 7) case both grow without bound. The spherical ( k = 1 ) ( k = 1 ) (k=1)(k=1)(k=1) case (Universe 6) turns over and re-collapses into a Big-Crunch singularity. The expansion factors for these universes are summarized in Fig. 18.4.

18.3 Einstein, Lemaître and Eddington

The universes so far have all had the potentially disturbing feature of being time dependent. The static Universe 8 is constructed such that it does not have this apparent problem. This model is known as the Einstein Universe. It contains dust and positive vacuum energy Λ E Λ E Lambda_(E)\Lambda_{\mathrm{E}}ΛE. We seek a static solution and see what happens.

Example 18.3

We start with the first Friedmann equation, written in the form for dust ρ d ( t ) ρ d ( t ) rho_(d)(t)\rho_{d}(t)ρd(t) and an amount of positive energy density Λ E Λ E Lambda_(E)\Lambda_{\mathrm{E}}ΛE
(18.21) a ˙ 2 + k = ( 8 π ρ d 3 + Λ E 3 ) a 2 (18.21) a ˙ 2 + k = 8 π ρ d 3 + Λ E 3 a 2 {:(18.21)a^(˙)^(2)+k=((8pirho_(d))/(3)+(Lambda_(E))/(3))a^(2):}\begin{equation*} \dot{a}^{2}+k=\left(\frac{8 \pi \rho_{\mathrm{d}}}{3}+\frac{\Lambda_{\mathrm{E}}}{3}\right) a^{2} \tag{18.21} \end{equation*}(18.21)a˙2+k=(8πρd3+ΛE3)a2
Static implies that a ˙ ( t ) = 0 , a ( t ) = a 0 a ˙ ( t ) = 0 , a ( t ) = a 0 a^(˙)(t)=0,a(t)=a_(0)\dot{a}(t)=0, a(t)=a_{0}a˙(t)=0,a(t)=a0 and ρ ( t ) = ρ d ( 0 ) ρ ( t ) = ρ d ( 0 ) rho(t)=rho_(d)^((0))\rho(t)=\rho_{\mathrm{d}}^{(0)}ρ(t)=ρd(0) and so the first Friedmann equation becomes
(18.22) k = ( 8 π 3 ρ d ( 0 ) + Λ E 3 ) a 0 2 (18.22) k = 8 π 3 ρ d ( 0 ) + Λ E 3 a 0 2 {:(18.22)k=((8pi)/(3)*rho_(d)^((0))+(Lambda_(E))/(3))a_(0)^(2):}\begin{equation*} k=\left(\frac{8 \pi}{3} \cdot \rho_{\mathrm{d}}^{(0)}+\frac{\Lambda_{\mathrm{E}}}{3}\right) a_{0}^{2} \tag{18.22} \end{equation*}(18.22)k=(8π3ρd(0)+ΛE3)a02
It's convenient to also make use of the second Friedmann equation here with a ¨ = a ˙ = a ¨ = a ˙ = a^(¨)=a^(˙)=\ddot{a}=\dot{a}=a¨=a˙= 0 , which then reads
(18.23) k a 0 2 = Λ E (18.23) k a 0 2 = Λ E {:(18.23)(k)/(a_(0)^(2))=Lambda_(E):}\begin{equation*} \frac{k}{a_{0}^{2}}=\Lambda_{\mathrm{E}} \tag{18.23} \end{equation*}(18.23)ka02=ΛE
From the positivity of Λ E Λ E Lambda_(E)\Lambda_{\mathrm{E}}ΛE we conclude that for a static universe we need to select a spatially spherical ( k = 1 ) ( k = 1 ) (k=1)(k=1)(k=1) universe since this is the only way to have a positive k k kkk. The second Friedmann equation then tells us that a 0 = Λ E 1 2 a 0 = Λ E 1 2 a_(0)=Lambda_(E)^(-(1)/(2))a_{0}=\Lambda_{\mathrm{E}}^{-\frac{1}{2}}a0=ΛE12 and substituting this into eqn 18.22 we obtain ρ d ( 0 ) = Λ E / 4 π ρ d ( 0 ) = Λ E / 4 π rho_(d)^((0))=Lambda_(E)//4pi\rho_{\mathrm{d}}^{(0)}=\Lambda_{\mathrm{E}} / 4 \piρd(0)=ΛE/4π. In words: a universe with dust density ρ d ( 0 ) ρ d ( 0 ) rho_(d)^((0))\rho_{\mathrm{d}}^{(0)}ρd(0) is kept static by the negative pressure of vacuum energy, provided that ρ d ( 0 ) = Λ E / 4 π ρ d ( 0 ) = Λ E / 4 π rho_(d)^((0))=Lambda_(E)//4pi\rho_{\mathrm{d}}^{(0)}=\Lambda_{\mathrm{E}} / 4 \piρd(0)=ΛE/4π. It follows that the radius of the universe is given in terms of the amount of dust matter by a 0 = ( 4 π ρ d ( 0 ) ) 1 2 a 0 = 4 π ρ d ( 0 ) 1 2 a_(0)=(4pirho_(d)^((0)))^(-(1)/(2))a_{0}=\left(4 \pi \rho_{\mathrm{d}}^{(0)}\right)^{-\frac{1}{2}}a0=(4πρd(0))12.
The static Einstein Universe (Universe 8) is shown in Fig. 18.7 where, because k = 1 k = 1 k=1k=1k=1, each spatial hypersurface represents a 3 -sphere. The Einstein Universe is static, but this doesn't mean it is stable. In fact, the next example shows that it cannot withstand a small perturbation in the density of the dust that fills it.
Example 18.4
We examine the stability of this universe 2 2 ^(2){ }^{2}2 by introducing a perturbation in both a ( t ) a ( t ) a(t)a(t)a(t) and ρ d ρ d rho_(d)\rho_{\mathrm{d}}ρd. Subtracting the first Friedmann equation from the second we obtain
(18.24) 2 a ¨ a = 8 π 3 ρ d + 2 3 Λ E (18.24) 2 a ¨ a = 8 π 3 ρ d + 2 3 Λ E {:(18.24)2((a^(¨)))/(a)=-(8pi)/(3)*rho_(d)+(2)/(3)Lambda_(E):}\begin{equation*} 2 \frac{\ddot{a}}{a}=-\frac{8 \pi}{3} \cdot \rho_{\mathrm{d}}+\frac{2}{3} \Lambda_{\mathrm{E}} \tag{18.24} \end{equation*}(18.24)2a¨a=8π3ρd+23ΛE
We perturb around the equilibrium values by setting a = Λ E 1 2 + δ a a = Λ E 1 2 + δ a a=Lambda_(E)^(-(1)/(2))+delta aa=\Lambda_{\mathrm{E}}^{-\frac{1}{2}}+\delta aa=ΛE12+δa and ρ d = ρ d = rho_(d)=\rho_{\mathrm{d}}=ρd= Λ E 4 π + δ ρ d Λ E 4 π + δ ρ d (Lambda_(E))/(4pi)+deltarho_(d)\frac{\Lambda_{\mathrm{E}}}{4 \pi}+\delta \rho_{\mathrm{d}}ΛE4π+δρd, giving, to order δ a δ a delta a\delta aδa, the equation of motion
(18.25) 2 Λ E 1 2 ( δ a ¨ ) = 8 π 3 δ ρ d (18.25) 2 Λ E 1 2 ( δ a ¨ ) = 8 π 3 δ ρ d {:(18.25)2Lambda_(E)^((1)/(2))(delta a^(¨))=-(8pi)/(3)deltarho_(d):}\begin{equation*} 2 \Lambda_{\mathrm{E}}^{\frac{1}{2}}(\ddot{\delta a})=-\frac{8 \pi}{3} \delta \rho_{\mathrm{d}} \tag{18.25} \end{equation*}(18.25)2ΛE12(δa¨)=8π3δρd
If the total amount of dust is constant, then we also have the thermodynamic result that
(18.26) ρ d a 3 = const., (18.26) ρ d a 3 =  const.,  {:(18.26)rho_(d)a^(3)=" const., ":}\begin{equation*} \rho_{\mathrm{d}} a^{3}=\text { const., } \tag{18.26} \end{equation*}(18.26)ρda3= const., 
and so we can find an equation for δ ρ d δ ρ d deltarho_(d)\delta \rho_{\mathrm{d}}δρd by differentiating:
(18.27) δ ρ d ρ d = 3 δ a a . (18.27) δ ρ d ρ d = 3 δ a a . {:(18.27)(deltarho_(d))/(rho_(d))=-3(delta a)/(a).:}\begin{equation*} \frac{\delta \rho_{\mathrm{d}}}{\rho_{\mathrm{d}}}=-3 \frac{\delta a}{a} . \tag{18.27} \end{equation*}(18.27)δρdρd=3δaa.
Substituting into eqn 18.25 , we conclude that
(18.28) ( δ a ¨ ) = Λ E ( δ a ) , (18.28) ( δ a ¨ ) = Λ E ( δ a ) , {:(18.28)(delta a^(¨))=Lambda_(E)(delta a)",":}\begin{equation*} (\ddot{\delta a})=\Lambda_{\mathrm{E}}(\delta a), \tag{18.28} \end{equation*}(18.28)(δa¨)=ΛE(δa),
whose solution is δ a e Λ E t δ a e Λ E t delta a prope^(sqrt(Lambda_(E)t))\delta a \propto \mathrm{e}^{\sqrt{\Lambda_{\mathrm{E}} t}}δaeΛEt. The result, therefore, is that the perturbation grows exponentially in time, causing the universe to evolve. The static universe is not in a stable equilibrium.
We conclude that the Einstein Universe is only just static, since it contains a critical amount of energy density Λ E = 4 π ρ d Λ E = 4 π ρ d Lambda_(E)=4pirho_(d)\Lambda_{\mathrm{E}}=4 \pi \rho_{\mathrm{d}}ΛE=4πρd. The vacuum energy tries to make the universe expand; the matter tends to cause the universe to contract and they end up in unstable equilibrium.
In Universe 9, known as the Lemaitre solution, 3 3 ^(3){ }^{3}3 we have k = 1 k = 1 k=1k=1k=1 and a universe full of dust and vacuum energy, but this time with Λ > Λ E Λ > Λ E Lambda > Lambda_(E)\Lambda>\Lambda_{\mathrm{E}}Λ>ΛE. Since the vacuum energy tends to make universes expand, we should expect that this universe expands without bound. However, there is a surprise: the model has a coasting period where it stays at roughly the same size. This coasting period can last for arbitrarily long time if the energy density is set appropriately. This is represented in Fig. 18.8, where each spatial hypersurface represents a 3 -sphere.
Fig. 18.7 The evolution of Universe 8, the Einstein Universe, which has k = k = k=k=k= 1 and vacuum energy Λ = Λ E Λ = Λ E Lambda=Lambda_(E)\Lambda=\Lambda_{\mathrm{E}}Λ=ΛE. Each circle represents a 3 -sphere.
2 2 ^(2){ }^{2}2 Here we follow the approach described
in the problem book by Lightman et al.
Fig. 18.8 The evolution of Universe 9 which has k = 1 k = 1 k=1k=1k=1 and vacuum energy Λ > Λ E Λ > Λ E Lambda > Lambda_(E)\Lambda>\Lambda_{\mathrm{E}}Λ>ΛE.

Example 18.5

Consider Universe 9 for the case where Λ = ( 1 + ε ) 2 Λ E Λ = ( 1 + ε ) 2 Λ E Lambda=(1+epsi)^(2)Lambda_(E)\Lambda=(1+\varepsilon)^{2} \Lambda_{\mathrm{E}}Λ=(1+ε)2ΛE, where ε ε epsi\varepsilonε is assumed small Since, for the Einstein Universe we have a 0 = ( 4 π ρ d ) 1 2 a 0 = 4 π ρ d 1 2 a_(0)=(4pirho_(d))^(-(1)/(2))a_{0}=\left(4 \pi \rho_{\mathrm{d}}\right)^{-\frac{1}{2}}a0=(4πρd)12 and a 0 = Λ E 1 2 a 0 = Λ E 1 2 a_(0)=Lambda_(E)^(-(1)/(2))a_{0}=\Lambda_{\mathrm{E}}^{-\frac{1}{2}}a0=ΛE12, we can write for the dust that
(18.29) 8 π ρ ( t ) 3 = 2 3 Λ E 1 2 a ( t ) 3 . (18.29) 8 π ρ ( t ) 3 = 2 3 Λ E 1 2 a ( t ) 3 . {:(18.29)(8pi rho(t))/(3)=(2)/(3Lambda_(E)^((1)/(2))a(t)^(3)).:}\begin{equation*} \frac{8 \pi \rho(t)}{3}=\frac{2}{3 \Lambda_{\mathrm{E}}^{\frac{1}{2}} a(t)^{3}} . \tag{18.29} \end{equation*}(18.29)8πρ(t)3=23ΛE12a(t)3.
Substituting Λ / ( 1 + ε ) 2 Λ / ( 1 + ε ) 2 Lambda//(1+epsi)^(2)\Lambda /(1+\varepsilon)^{2}Λ/(1+ε)2 for Λ E Λ E Lambda_(E)\Lambda_{\mathrm{E}}ΛE and setting k = 1 k = 1 k=1k=1k=1, we write the first Friedmann equation in the form
(18.30) a ˙ 2 + 1 = V eff ( a ) (18.30) a ˙ 2 + 1 = V eff ( a ) {:(18.30)a^(˙)^(2)+1=V_(eff)(a):}\begin{equation*} \dot{a}^{2}+1=V_{\mathrm{eff}}(a) \tag{18.30} \end{equation*}(18.30)a˙2+1=Veff(a)
where we've defined an effective potential
(18.31) V eff ( a ) = 2 ( 1 + ε ) 3 Λ 1 2 a + Λ 3 a 2 (18.31) V eff ( a ) = 2 ( 1 + ε ) 3 Λ 1 2 a + Λ 3 a 2 {:(18.31)V_(eff)(a)=(2(1+epsi))/(3Lambda^((1)/(2))a)+(Lambda)/(3)a^(2):}\begin{equation*} V_{\mathrm{eff}}(a)=\frac{2(1+\varepsilon)}{3 \Lambda^{\frac{1}{2}} a}+\frac{\Lambda}{3} a^{2} \tag{18.31} \end{equation*}(18.31)Veff(a)=2(1+ε)3Λ12a+Λ3a2
From the last example we see that when a a aaa is small, the radius of Universe 9 varies as a ˙ 1 / a 1 2 a ˙ 1 / a 1 2 a^(˙)prop1//a^((1)/(2))\dot{a} \propto 1 / a^{\frac{1}{2}}a˙1/a12. We examined this behaviour in the Einstein-de Sitter model (Universe 3) and found that it implies that the system expands from an initial singularity as a t 2 3 a t 2 3 a propt^((2)/(3))a \propto t^{\frac{2}{3}}at23. However, as a a aaa becomes larger the rate of expansion slows, reaching a minimum when a ˙ a ˙ a^(˙)\dot{a}a˙ is a minimum. This is the case when 4 4 ^(4){ }^{4}4
(18.32) a = a min = ( 1 + ε ) 1 3 Λ 1 2 (18.32) a = a min = ( 1 + ε ) 1 3 Λ 1 2 {:(18.32)a=a_(min)=((1+epsi)^((1)/(3)))/(Lambda^((1)/(2))):}\begin{equation*} a=a_{\min }=\frac{(1+\varepsilon)^{\frac{1}{3}}}{\Lambda^{\frac{1}{2}}} \tag{18.32} \end{equation*}(18.32)a=amin=(1+ε)13Λ12
The universe then coasts for a period, before speeding up again.
Example 18.6
Expanding the effective potential V ( a ) V ( a ) V(a)V(a)V(a) about the minimum, we have
a ˙ 2 1 + V ( a min ) + 1 2 ( a a min ) 2 2 V a 2 | a = a min (18.33) = 1 + ( 1 + ε ) 2 3 + [ Λ 1 2 a ( 1 + ε ) 1 3 ] 2 a ˙ 2 1 + V a min + 1 2 a a min 2 2 V a 2 a = a min (18.33) = 1 + ( 1 + ε ) 2 3 + Λ 1 2 a ( 1 + ε ) 1 3 2 {:[a^(˙)^(2)~~-1+V(a_(min))+(1)/(2)(a-a_(min))^(2)(del^(2)V)/(dela^(2))|_(a=a_(min))],[(18.33)=-1+(1+epsi)^((2)/(3))+[Lambda^((1)/(2))a-(1+epsi)^((1)/(3))]^(2)]:}\begin{align*} \dot{a}^{2} & \approx-1+V\left(a_{\min }\right)+\left.\frac{1}{2}\left(a-a_{\min }\right)^{2} \frac{\partial^{2} V}{\partial a^{2}}\right|_{a=a_{\min }} \\ & =-1+(1+\varepsilon)^{\frac{2}{3}}+\left[\Lambda^{\frac{1}{2}} a-(1+\varepsilon)^{\frac{1}{3}}\right]^{2} \tag{18.33} \end{align*}a˙21+V(amin)+12(aamin)22Va2|a=amin(18.33)=1+(1+ε)23+[Λ12a(1+ε)13]2
This equation can be integrated. The result (a standard integral that can be looked up) is
a ( t ) = ( 1 + ε ) 1 3 Λ 1 2 { 1 + [ 1 ( 1 + ε ) 2 3 ] 1 2 sinh [ Λ 1 2 ( t t m ) ] } a ( t ) = ( 1 + ε ) 1 3 Λ 1 2 1 + 1 ( 1 + ε ) 2 3 1 2 sinh Λ 1 2 t t m a(t)=((1+epsi)^((1)/(3)))/(Lambda^((1)/(2))){1+[1-(1+epsi)^(-(2)/(3))]^((1)/(2))sinh[Lambda^((1)/(2))(t-t_(m))]}a(t)=\frac{(1+\varepsilon)^{\frac{1}{3}}}{\Lambda^{\frac{1}{2}}}\left\{1+\left[1-(1+\varepsilon)^{-\frac{2}{3}}\right]^{\frac{1}{2}} \sinh \left[\Lambda^{\frac{1}{2}}\left(t-t_{\mathrm{m}}\right)\right]\right\}a(t)=(1+ε)13Λ12{1+[1(1+ε)23]12sinh[Λ12(ttm)]}
where t m t m t_(m)t_{\mathrm{m}}tm is the time at which a ˙ a ˙ a^(˙)\dot{a}a˙ reaches its minimum. In the limit of small ε ε epsi\varepsilonε, we
Fig. 18.9 The radius function a ( t ) a ( t ) a(t)a(t)a(t) for the some k = 1 k = 1 k=1k=1k=1 universes. (a) Universe 9 (the Lemaître model) with Λ > Λ E Λ > Λ E Lambda > Lambda_(E)\Lambda>\Lambda_{\mathrm{E}}Λ>ΛE showing its famous coasting period; (b) Universe 8 (the Einstein static universe with Λ E Λ E Lambda_(E)\Lambda_{\mathrm{E}}ΛE ) and Universe 10 (The Eddington-Lemaitre model), 10 (The Eddington-Lemaitre model),
in its two forms, distinguished by different initial conditions; (c) Universe 11, with 0 < Λ < Λ E 0 < Λ < Λ E 0 < Lambda < Lambda_(E)0<\Lambda<\Lambda_{\mathrm{E}}0<Λ<ΛE for two distinct initial conditions.
have a a min [ 1 + ε 3 sinh Λ 1 2 ( t t m ) ] a a min 1 + ε 3 sinh Λ 1 2 t t m quad a~~a_(min)[1+(epsi)/(3)*sinh Lambda^((1)/(2))(t-t_(m))]\quad a \approx a_{\min }\left[1+\frac{\varepsilon}{3} \cdot \sinh \Lambda^{\frac{1}{2}}\left(t-t_{\mathrm{m}}\right)\right]aamin[1+ε3sinhΛ12(ttm)].
As a result, a a aaa is approximately a min a min a_(min)a_{\min }amin until
ε sinh Λ 1 2 ( t t m ) 1 ε sinh Λ 1 2 t t m 1 epsi sinh Lambda^((1)/(2))(t-t_(m))~~1\varepsilon \sinh \Lambda^{\frac{1}{2}}\left(t-t_{\mathrm{m}}\right) \approx 1εsinhΛ12(ttm)1
which is to say, for a time
(18.36) ( t t m ) ln ( 1 ε ) Λ 1 2 (18.36) t t m ln 1 ε Λ 1 2 {:(18.36)(t-t_(m))~~(ln((1)/(epsi)))/(Lambda^((1)/(2))):}\begin{equation*} \left(t-t_{\mathrm{m}}\right) \approx \frac{\ln \left(\frac{1}{\varepsilon}\right)}{\Lambda^{\frac{1}{2}}} \tag{18.36} \end{equation*}(18.36)(ttm)ln(1ε)Λ12
This coasting period can therefore be made arbitrarily long with a sufficiently small ε ε epsi\varepsilonε. Eventually, the expansion starts again with ( a ˙ / a ) 2 Λ / 3 ( a ˙ / a ) 2 Λ / 3 (a^(˙)//a)^(2)rarr Lambda//3(\dot{a} / a)^{2} \rightarrow \Lambda / 3(a˙/a)2Λ/3 as we had for the de Sitter model (Universe 1).
18.3 Einstein, Lemaître and Eddington 195
Fig. 18.10 The origins and fates of some Robertson-Walker spacetimes. These were computed for dust and vacuum energy universes using eqn 18.11.
5 5 ^(5){ }^{5}5 Arthur Eddington (1882-1944) was an early champion of general relativity, writing several articles describing it for an English-speaking audience, who were cut off from many developments in Germany owing to World War I.
Fig. 18.11 Universe 10, the Eddington-Lemaître model, starts similarly to the Einstein static universe and begins to expand after a certain time.
Fig. 18.12 Universe 11, showing an initial contraction, followed by a reexpansion.
The behaviour of the radius function in Universe 9 is shown in Fig. 18.9(a).
As a result of studying the Lemaitre model, Arthur Eddington 5 5 ^(5){ }^{5}5 proposed a model without a Big-Bang singularity. Instead the model starts by coasting at a value of a a aaa close to the static Einstein value until, at some point, it starts expanding. This is Universe 10: the EddingtonLemaître model shown in Fig. 18.9(b) and Fig. 18.11. Actually, there is another variant of this model: we could start from a Big-Bang singularity and asymptotically approach the Einstein Universe as shown in Fig. 18.9(b).
Finally, what if we have 0 < Λ < Λ E 0 < Λ < Λ E 0 < Lambda < Lambda_(E)0<\Lambda<\Lambda_{\mathrm{E}}0<Λ<ΛE ? This is the content of Universe 11. In this case, there are two possibilities, depending on the initial conditions. If we start from an initial Big-Bang singularity then there will not be enough vacuum energy to prevent the matter collapsing back in itself and we have a Big Crunch singularity. Perhaps more interestingly, if we start at a very large value of a a aaa, then the universe contracts, reaches a minimum radius, and then expands again. It has a e H ( t ) a e H ( t ) a prope^(H(-t))a \propto \mathrm{e}^{H(-t)}aeH(t) as t t t rarr-oot \rightarrow-\inftyt and a e H t a e H t a prope^(Ht)a \propto \mathrm{e}^{H t}aeHt as t t t rarr oot \rightarrow \inftyt. This is shown in Fig. 18.12 with a radius function as shown in Fig. 18.9(c).
We can summarize some of our progress by examining the evolution of several universes in Fig. 18.10.

18.4 A brief history of model universes

We have presented several model universes generated by varying the content and geometry of the models described by the Friedmann equations. This was not the order in which they were discussed historically. The discovery that our Universe is expanding was established through observations made by a number of people in the decades following 1910. Perhaps most famous of the astronomers in this period is Edwin Hubble, who in 1929 established that the speed of recession of a galaxy v d v d v_(d)v_{\mathrm{d}}vd is proportional to its distance from us. This is Hubble's law that v d = H × v d = H × v_(d)=H xxv_{\mathrm{d}}=H \timesvd=H× distance.
Somewhat earlier, back in 1917, Einstein had considered filling a spatially spherical universe with uniform matter and realized that, owing to the gravitational attraction of the matter, the resulting model could not be static without the incorporation into the theory of an extra energy contribution Λ Λ Lambda\LambdaΛ, that he called the cosmological term. In the same year, de Sitter worked out that, for an isotropic model, the cosmologicalconstant term could provide a repulsion that exactly cancelled the gravitational attraction. This gives us the Einstein model (Universe 8). The model ceased to be considered seriously after it became established that the physical Universe was expanding. Eddington in 1930 showed that this universe was unstable to perturbation and, in 1931, Einstein recommended the cosmological term be dropped from general relativity.
In 1917, de Sitter had also observed that he could model an expanding universe if he included only the Λ Λ Lambda\LambdaΛ term. Alexander Friedmann derived his results for evolving isotropic and homogenous universes in 1922, which also suggested the Hubble-law expansion v d = H × v d = H × v_(d)=H xxv_{\mathrm{d}}=H \timesvd=H× distance. (In 1923, Hermann Weyl pointed out that test particles in the de Sitter model would separate at a rate consistent with this law.) In 1927, the Catholic priest and physicist George Lemaître independently derived Friedmann's homogeneous, isotropic models and proposed the Eddington-Lemaître model (Universe 10). This suggestion of a model that started at the Einstein solution and expanded was promoted by Eddington. However, Lemaitre would in 1931 suggest that the universe could have expanded from what he termed a 'Primeval Atom' (his Universe 9), an idea later described rather dismissively by Fred Hoyle 6 6 ^(6){ }^{6}6 as the 'Big Bang theory'. Robertson and Walker would independently show in 1935 that the line element Lemaître was using represents the most general geometry compatible with homogeneity and isotropy.
There was a significant problem with models involving the expansion of the universe from an initial point in spacetime. The age of rocks (and hence the Earth) had been estimated from the radioactive properties of uranium to be at least 3 × 10 9 3 × 10 9 3xx10^(9)3 \times 10^{9}3×109 years, but this didn't seem to allow long enough for the universe to evolve into its current state.

Example 18.7

Let's start by assuming that the acceleration of our Universe has always been negative (i.e. the rate of expansion has always been slowing down). If we consider the evolution of a ( t ) a ( t ) a(t)a(t)a(t) then its tangent evaluated now a ˙ ( t now ) a ˙ t now  a^(˙)(t_("now "))\dot{a}\left(t_{\text {now }}\right)a˙(tnow ) cuts the axis at an earlier time t 1 t 1 t_(1)t_{1}t1 where
(18.38) a ( t now ) = a ˙ ( t now ) ( t now t 1 ) (18.38) a t now  = a ˙ t now  t now  t 1 {:(18.38)a(t_("now "))=a^(˙)(t_("now "))(t_("now ")-t_(1)):}\begin{equation*} a\left(t_{\text {now }}\right)=\dot{a}\left(t_{\text {now }}\right)\left(t_{\text {now }}-t_{1}\right) \tag{18.38} \end{equation*}(18.38)a(tnow )=a˙(tnow )(tnow t1)
This is shown in Fig. 18.13. Rearranging, we find
(18.39) t now t 1 = a ( t now ) a ˙ ( t now ) = H ( t now ) 1 (18.39) t now  t 1 = a t now  a ˙ t now  = H t now  1 {:(18.39)t_("now ")-t_(1)=(a(t_("now ")))/((a^(˙))(t_("now ")))=H(t_("now "))^(-1):}\begin{equation*} t_{\text {now }}-t_{1}=\frac{a\left(t_{\text {now }}\right)}{\dot{a}\left(t_{\text {now }}\right)}=H\left(t_{\text {now }}\right)^{-1} \tag{18.39} \end{equation*}(18.39)tnow t1=a(tnow )a˙(tnow )=H(tnow )1
Before the 1950 s the quantity H ( t now ) 1 H t now  1 H(t_("now "))^(-1)H\left(t_{\text {now }}\right)^{-1}H(tnow )1 was estimated to be 1.8 × 10 9 1.8 × 10 9 1.8 xx10^(9)1.8 \times 10^{9}1.8×109 years. If the universe does start at, or around, a ( t = 0 ) = 0 a ( t = 0 ) = 0 a(t=0)=0a(t=0)=0a(t=0)=0, then its age must be less than H ( t now ) 1 H t now  1 H(t_("now "))^(-1)H\left(t_{\text {now }}\right)^{-1}H(tnow )1.
If the ages of the rocks are as claimed on the basis of the nuclear physics, then there doesn't seem to have been enough time for the Universe to evolve before the creation of rocks. This is known as the timescale problem. One solution was the introduction into the Lemaitre model of large positive values of Λ Λ Lambda\LambdaΛ that cause a large acceleration. It is also solved by the Eddington-Lemaître model, since this starts from the Einstein solution before expansion. However, in that model the assumption that the stars condense at the moment that the expansion starts is problematic, since it is not clear how this can happen within a static and homogeneous model.
By the 1930 s, nuclear physics was also offering explanations of the genesis of the elements (nucleogenesis) which seemed consistent with Big
6 6 ^(6){ }^{6}6 Fred Hoyle (1915-2001) rejected the Big Bang theory and supported an alternative steady-state picture, in which there is no cosmological constant, but where a creation field (or C-field) is responsible for matter generation as the Universe expands, allowing the density to remain constant in time. The C-field has negative energy density (allowing for conservation of energy as new matter is created) and so exerts a negative pressure. The inability of the steadypressur. Thel to account for state model to acound fise cosmic microwave background, discovered in 1964, is one of the reasons why it was abandoned by cosmologists. The term 'Big Bang' was coined in a talk given by Hoyle on BBC radio in 1949. Hoyle denied that he intended to be insulting; rather that he was emphasizing the difference between the models for a general audience.
Fig. 18.13 Estimation of the age T T TTT of the Universe using a ( t ) a ( t ) a(t)a(t)a(t). Assuming the Universe has been monotonically decelerating, we must have T < t now t 1 T < t now  t 1 T < t_("now ")-t_(1)T<t_{\text {now }}-t_{1}T<tnow t1.
Fig. 18.14 Selected measurements of H 0 H 0 H_(0)H_{0}H0.
\curvearrowright The evolution of the very early Universe is discussed in Chapter 41. See Chapter 15 for a brief description of the anisotropy of the CMB.
7 7 ^(7){ }^{7}7 Recent estimates give h h hhh to be close to 0.7 (see Fig. 18.14).
\curvearrowright In Section 49.8 we will discuss our current understanding, based on experimental evidence, on the question of which type of Universe we are actually living in.
Bang models, since as the scale factor a a aaa tends to zero, the temperature should go as T a 1 T a 1 T propa^(-1)T \propto a^{-1}Ta1 allowing, in the early moments after the Big Bang, the extreme conditions needed to fuse hydrogen to make the heavier elements. This led Lemaître to promote the idea of a hot Big Bang.
By the 1950s new measurements of Hubble's constant H 0 H 0 H_(0)H_{0}H0 (see Fig. 18.14 for a timeline) suggested an increased lifetime of the Universe of 1.3 × 10 10 1.3 × 10 10 1.3 xx10^(10)1.3 \times 10^{10}1.3×1010 years, effectively resolving the timescale problem without the need for a large cosmological constant, so that Big Bang models with Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0 could now be seriously considered. This had an aesthetic appeal, owing to the slightly ad-hoc nature of the introduction of the cosmological constant. The Λ = 0 Λ = 0 Lambda=0\Lambda=0Λ=0 models became known as the standard models of cosmology. These are, for k = 0 k = 0 k=0k=0k=0, Universe 3 (Einstein-de Sitter), for k = 1 k = 1 k=1k=1k=1, Universe 6 and for k = 1 k = 1 k=-1k=-1k=1, Universe 7. They are all Big Bang models.
Compelling evidence for a Big-Bang cosmology came from the measurement of the cosmic microwave background (CMB), which is the faint electromagnetic radiation remaining from recombination events that occurred during the early stages of the Universe. The existence of the CMB had originally been predicted in 1948 by Ralph Alpher and Robert Herman, following work by George Gamow. Alpher and Herman predicted that high-energy radiation from the very early Universe should have been shifted into the microwave region of the electromagnetic spectrum, and that this should result in an effective black-body temperature of space of T = 5 K T = 5 K T=5KT=5 \mathrm{~K}T=5 K. The CMB was accidentally discovered by Arno Penzias and Robert Wilson in 1964, and has since been subject to very precise measurements, that determine its temperatures to be 2.73 K , with faint anisotropies.
Which of the three standard models our Universe most resembles is determined by how the mass-energy content of the Universe compares with the critical density (discussed in the last chapter) of ρ c = 3 H 0 2 / 8 π ρ c = 3 H 0 2 / 8 π rho_(c)=3H_(0)^(2)//8pi\rho_{\mathrm{c}}=3 H_{0}^{2} / 8 \piρc=3H02/8π. Expressing the Hubble constant 7 7 ^(7){ }^{7}7 as H 0 = h × 100 km s 1 Mpc 1 H 0 = h × 100 km s 1 Mpc 1 H_(0)=h xx100kms^(-1)Mpc^(-1)H_{0}=h \times 100 \mathrm{~km} \mathrm{~s}^{-1} \mathrm{Mpc}^{-1}H0=h×100 km s1Mpc1, the critical density is ρ c = 1.88 × 10 26 h 2 kg m 3 ρ c = 1.88 × 10 26 h 2 kg m 3 rho_(c)=1.88 xx10^(-26)h^(2)kgm^(-3)\rho_{\mathrm{c}}=1.88 \times 10^{-26} h^{2} \mathrm{~kg} \mathrm{~m}^{-3}ρc=1.88×1026h2 kg m3. Observations suggests that the density of observable matter in the Universe is actually only a small fraction of this, at around 10 28 kg m 3 10 28 kg m 3 10^(-28)kgm^(-3)10^{-28} \mathrm{~kg} \mathrm{~m}^{-3}1028 kg m3. However, observations also suggest that the properties of the Universe are consistent with a flat, k = 0 k = 0 k=0k=0k=0 model. This discrepancy is known as the missing mass problem and remains one of the key questions about our knowledge of the Universe.
So far we have been quite cavalier in accepting the presence of the Big-Bang singularity. We are now at the point where we must engage with it, and the break down of physical laws that it brings. This is the goal of the next chapter.

Chapter summary

  • The Robertson-Walker spaces filled with a perfect fluid provide a range of models, many of which start with a Big Bang.
  • The Einstein Universe is static, but unstable; the Einstein-de Sitter model expands without bound and the Lemaitre model has a coasting period.
  • The standard models of cosmology have zero cosmological constant Λ Λ Lambda\LambdaΛ, and all begin with a Big-Bang singularity.

Exercises

18.1) Let's examine a Robertson-Walker universe with no matter or vacuum energy in it, by letting ρ = p = Λ = 0 ρ = p = Λ = 0 rho=p=Lambda=0\rho=p=\Lambda=0ρ=p=Λ=0
(a) Set k = 1 k = 1 k=-1k=-1k=1 and show that the resulting metric line element can be written as
d s 2 = d t 2 + t 2 [ d χ 2 + sinh 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] d s 2 = d t 2 + t 2 d χ 2 + sinh 2 χ d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-dt^(2)+t^(2)[(d)chi^(2)+sinh^(2)chi((d)theta^(2)+sin^(2)theta(d)phi^(2))]\mathrm{d} s^{2}=-\mathrm{d} t^{2}+t^{2}\left[\mathrm{~d} \chi^{2}+\sinh ^{2} \chi\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right]ds2=dt2+t2[ dχ2+sinh2χ( dθ2+sin2θ dϕ2)].
(18.40)
(b) Using the transformations
(18.41) r = t sinh χ , T = t cosh χ (18.41) r = t sinh χ , T = t cosh χ {:(18.41)r=t sinh chi","quad T=t cosh chi:}\begin{equation*} r=t \sinh \chi, \quad T=t \cosh \chi \tag{18.41} \end{equation*}(18.41)r=tsinhχ,T=tcoshχ
show that this is a familiar spacetime in disguise.
(18.2) Verify by substitution that (a) eqn 18.13 and (b) eqn 18.14 are solutions to the Friedmann equations.
(18.3) Verify eqns 18.25 and 18.28 .
(18.4) Verify eqns 18.32 and 18.33 .
There is a neat geometrical interpretation of the de Sitter universes, which are those that are driven only by a non-zero cosmological constant A. (Universe 1 is the k = 0 k = 0 k=0k=0k=0 version. There are also k = ± 1 k = ± 1 k=+-1k= \pm 1k=±1 versions.) The construction involves embedding a hyperboloid
T 2 + W 2 + X 2 + Y 2 + Z 2 = α 2 T 2 + W 2 + X 2 + Y 2 + Z 2 = α 2 -T^(2)+W^(2)+X^(2)+Y^(2)+Z^(2)=alpha^(2)-T^{2}+W^{2}+X^{2}+Y^{2}+Z^{2}=\alpha^{2}T2+W2+X2+Y2+Z2=α2
with α α alpha\alphaα a constant, in a ( 4 + 1 ) a ( 4 + 1 ) a(4+1)a(4+1)a(4+1)-dimensional Minkowski space with line element d s 2 = d T 2 + d s 2 = d T 2 + ds^(2)=-dT^(2)+\mathrm{d} s^{2}=-\mathrm{d} T^{2}+ds2=dT2+ d X 2 + d Y 2 + d Z 2 + d W 2 d X 2 + d Y 2 + d Z 2 + d W 2 dX^(2)+dY^(2)+dZ^(2)+dW^(2)\mathrm{d} X^{2}+\mathrm{d} Y^{2}+\mathrm{d} Z^{2}+\mathrm{d} W^{2}dX2+dY2+dZ2+dW2. The hyperboloid is shown in Fig. 18.15, with two of the dimensions suppressed (i.e. the circular cuts should be regarded as 3-spheres in five dimensions). This seems very
abstract, by the idea of covering the hyperboloid with different coordinate system to see the features of the physics, can be demonstrated on the two-dimensional surface of a hyperboloid described by T 2 + X 2 + Y 2 = 1 T 2 + X 2 + Y 2 = 1 -T^(2)+X^(2)+Y^(2)=1-T^{2}+X^{2}+Y^{2}=1T2+X2+Y2=1, embedded in ( 2 + 1 ) ( 2 + 1 ) (2+1)(2+1)(2+1) dimensional Minkowski space, as demonstrated by the following exercises.
Fig. 1 8 . 1 5 1 8 . 1 5 18.15\mathbf{1 8 . 1 5}18.15 de Sitter spacetime represented as a hyperboloid.
(18.5) As a warm-up exercise, use the method from Appendix D to embed the hyperbola T 2 + X 2 = 1 T 2 + X 2 = 1 -T^(2)+X^(2)=1-T^{2}+X^{2}=1T2+X2=1 in (1+1)-dimensional Minkowski space with line element d s 2 = d T 2 + d X 2 d s 2 = d T 2 + d X 2 ds^(2)=-dT^(2)+dX^(2)\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} X^{2}ds2=dT2+dX2.
(18.6) Embed the hyperbolic surface T 2 + X 2 + Y 2 = 1 T 2 + X 2 + Y 2 = 1 -T^(2)+X^(2)+Y^(2)=1-T^{2}+X^{2}+Y^{2}=1T2+X2+Y2=1 shown in Fig. 18.15 in Minkowski space with line element d s 2 = d T 2 + d X 2 + d Y 2 d s 2 = d T 2 + d X 2 + d Y 2 ds^(2)=-dT^(2)+dX^(2)+dY^(2)\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} X^{2}+\mathrm{d} Y^{2}ds2=dT2+dX2+dY2 by eliminating the variable Y Y YYY, to show
(a) the line element can be written in terms of variables v v vvv and ψ ψ psi\psiψ as
(18.43) d s 2 = d v 2 1 + v 2 + v 2 d ψ 2 (18.43) d s 2 = d v 2 1 + v 2 + v 2 d ψ 2 {:(18.43)ds^(2)=-(dv^(2))/(1+v^(2))+v^(2)dpsi^(2):}\begin{equation*} \mathrm{d} s^{2}=-\frac{\mathrm{d} v^{2}}{1+v^{2}}+v^{2} \mathrm{~d} \psi^{2} \tag{18.43} \end{equation*}(18.43)ds2=dv21+v2+v2 dψ2
(b) the latter expression can be reduced to the hyperbolic line element
(18.44) d s 2 = d χ 2 + sinh 2 χ d ψ 2 (18.44) d s 2 = d χ 2 + sinh 2 χ d ψ 2 {:(18.44)ds^(2)=-dchi^(2)+sinh^(2)chidpsi^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} \chi^{2}+\sinh ^{2} \chi \mathrm{~d} \psi^{2} \tag{18.44} \end{equation*}(18.44)ds2=dχ2+sinh2χ dψ2
where you should determine the relationship between coordinates ( χ , ψ ) ( χ , ψ ) (chi,psi)(\chi, \psi)(χ,ψ) and ( T , X , Y ) ( T , X , Y ) (T,X,Y)(T, X, Y)(T,X,Y).
(c) Compare the latter with the other form of the line element from the chapter, which was
(18.45) d s 2 = d χ 2 + cosh 2 χ d θ 2 (18.45) d s 2 = d χ 2 + cosh 2 χ d θ 2 {:(18.45)ds^(2)=-dchi^(2)+cosh^(2)chidtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} \chi^{2}+\cosh ^{2} \chi \mathrm{~d} \theta^{2} \tag{18.45} \end{equation*}(18.45)ds2=dχ2+cosh2χ dθ2
At constant timelike variable χ χ chi\chiχ there is a close resemblance between (i) eqn 18.44 and the line element of a spherical universe with a ( t ) = sinh t a ( t ) = sinh t a(t)=sinh ta(t)=\sinh ta(t)=sinht and (ii) eqn 18.45 and the line element of a hyperbolic Universe with a ( t ) = cosh t a ( t ) = cosh t a(t)=cosh ta(t)=\cosh ta(t)=cosht.
(d) Confirm this is the case by considering the Friedmann equations.
Notice how that, by choosing coordinates that split the space and time up in different ways, we are able to use de Sitter spacetime to represent cosmological-constant-driven models with different spatial curvatures. This works because the different cross sections of the hyperboloid represented by each coordinate system possess different spatial curvatures.
(18.7) Start again with a hyperbolic surface T 2 + T 2 + -T^(2)+-T^{2}+T2+ X 2 + Y 2 = 1 X 2 + Y 2 = 1 X^(2)+Y^(2)=1X^{2}+Y^{2}=1X2+Y2=1 embedded in (2+1)-dimensional Minkowski space in the form that results from eliminating Y Y YYY :
(18.46) d s 2 = d T 2 + d X 2 + ( T d T X d X ) 2 ( 1 + T 2 X 2 ) (18.46) d s 2 = d T 2 + d X 2 + ( T d T X d X ) 2 1 + T 2 X 2 {:(18.46)ds^(2)=-dT^(2)+dX^(2)+((T(d)T-X(d)X)^(2))/((1+T^(2)-X^(2))):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} X^{2}+\frac{(T \mathrm{~d} T-X \mathrm{~d} X)^{2}}{\left(1+T^{2}-X^{2}\right)} \tag{18.46} \end{equation*}(18.46)ds2=dT2+dX2+(T dTX dX)2(1+T2X2)
Now define variables
T = sinh t + x 2 2 e t (18.47) X = cosh t x 2 2 e t T = sinh t + x 2 2 e t (18.47) X = cosh t x 2 2 e t {:[T=sinh t+(x^(2))/(2)e^(t)],[(18.47)X=cosh t-(x^(2))/(2)e^(t)]:}\begin{align*} & T=\sinh t+\frac{x^{2}}{2} \mathrm{e}^{t} \\ & X=\cosh t-\frac{x^{2}}{2} \mathrm{e}^{t} \tag{18.47} \end{align*}T=sinht+x22et(18.47)X=coshtx22et
implying Y = x e t Y = x e t Y=xe^(t)Y=x \mathrm{e}^{t}Y=xet.
(a) Show that the line element becomes
(18.48) d s 2 = d t 2 + e 2 t d x 2 (18.48) d s 2 = d t 2 + e 2 t d x 2 {:(18.48)ds^(2)=-dt^(2)+e^(2t)dx^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{e}^{2 t} \mathrm{~d} x^{2} \tag{18.48} \end{equation*}(18.48)ds2=dt2+e2t dx2
which is the line element for the spatially expanding de Sitter Universe (Universe 1) in (1+1) dimensions.
(b) Show that these coordinates only cover half of the hyperboloid.
(c) What kind of fixed-time slices do these coordinates make on the hyperboloid?
(d) Find a transformation that turns eqn 18.48 into the line element
(18.49) d s 2 = 1 u 2 ( d u 2 + d x 2 ) (18.49) d s 2 = 1 u 2 d u 2 + d x 2 {:(18.49)ds^(2)=(1)/(u^(2))(-du^(2)+dx^(2)):}\begin{equation*} \mathrm{d} s^{2}=\frac{1}{u^{2}}\left(-\mathrm{d} u^{2}+\mathrm{d} x^{2}\right) \tag{18.49} \end{equation*}(18.49)ds2=1u2(du2+dx2)
The latter is a type of Poincaré line element for Minkowski space, demonstrating that the Poincaré line element describes a hyperbolic spacetime.
(18.8) Using the same method as in Exercise 18.7, show that a description of (1+1)-dimensional de Sitter spacetime using coordinates
T = ( 1 x ) 1 2 sinh t (18.50) X = ( 1 x 2 ) 1 2 cosh t T = ( 1 x ) 1 2 sinh t (18.50) X = 1 x 2 1 2 cosh t {:[T=(1-x)^((1)/(2))sinh t],[(18.50)X=(1-x^(2))^((1)/(2))cosh t]:}\begin{align*} & T=(1-x)^{\frac{1}{2}} \sinh t \\ & X=\left(1-x^{2}\right)^{\frac{1}{2}} \cosh t \tag{18.50} \end{align*}T=(1x)12sinht(18.50)X=(1x2)12cosht
gives the Schwarzschild-like static line element
(18.51) d s 2 = ( 1 x 2 ) d t 2 + d x 2 1 x 2 (18.51) d s 2 = 1 x 2 d t 2 + d x 2 1 x 2 {:(18.51)ds^(2)=-(1-x^(2))dt^(2)+(dx^(2))/(1-x^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-x^{2}\right) \mathrm{d} t^{2}+\frac{\mathrm{d} x^{2}}{1-x^{2}} \tag{18.51} \end{equation*}(18.51)ds2=(1x2)dt2+dx21x2
How much of the hyperboloid is covered by these coordinates?
The key feature of this coordinate system is that it is static. Since a cosmological constant is timeindependent, this is not really a surprise.

Causality, infinity, and horizons

19.1 Penrose diagrams 202
19.2 The de Sitter spacetime 209 19.3 Big-Bang singularities 211 Chapter summary 214
Exercises 214
Fig. 19.1 (a) A light cone in Minkowski space distinguishes past from future. (b) In an orientable spacetime (here think of wrapping this sheet of spacetime around into a cylinder by identifying the left and right sides), we can uniquely identify future-pointing can uniquely identify future-pointing
and past-pointing directions. (c) In a and past-pointing directions. (c) In a non-orientable spacetime, there is no way of uniquely determining a future direction and a past direction.
1 1 ^(1){ }^{1}1 We shall meet some, for example: in the interior of certain types of black hole.
Fig. 19.2 Closed timelike curves.
2 2 ^(2){ }^{2}2 Roger Penrose (1931- ). His influence on modern general relativity is difficult to overstate, including contributions to the theory of singularities and the physics of black holes for which he was awarded the Nobel Prize in Physics in 2020.
3 3 ^(3){ }^{3}3 Conformal transformations are those that locally preserve angles.
คTopology, the study of those properties of shapes or spaces that are preserved despite deformations, is introduced in Appendix C.
4 4 ^(4){ }^{4}4 These are sometimes known as Penrose-Carter diagrams, also crediting Brandon Carter (1942- ). Carter also contributed to the proof of the nohair theorem described in Chapter 29, and named the anthropic principle. The latter was explained by Penros The latter was explained by Penrose as follows: 'The argument can be use to explain why the conditions happen to be just right for the existence
of (intelligent) life on the Earth at the present time. For if they were not just right, then we should not have found ourselves to be here now, but somewhere else, at some other appropriate time.'
ancestors). Fortunately, there are very few regions of spacetime that can support timelike loops; 1 1 ^(1){ }^{1}1 instead, timelike curves are generally open, joining events far in the past to events far in the future, allowing us to distinguish past and future owing to the orientation of the light cones along the curve.
In addition to causality, when we investigate which events can possibly occur, we are also forced to understand what we mean by infinity: the limiting behaviour over large intervals of spacetime. What does it mean for a signal to come from very far, or infinitely far, away? If we wait long enough, can we potentially receive signals from the entire universe? Do we have a unlimited realm of influence or, even if we wait forever, are there regions of the universe we cannot, even in principle, contact with a light signal? In this and future chapters, it will become clear that the answers to these questions depend on the large-scale structure of spacetime and, in general, our domain of influence as observers is limited. Just as the horizon on the skyline marks the boundary between those parts of the world we can see and those we cannot, we shall identify horizons that divide events in spacetime into those that observers can access and those they cannot.
One thing we have largely neglected so far in our discussion of the model universes is the presence and nature of infinities. In this chapter, we introduce a toolkit for understanding infinity that was developed in the 1960 s, largely by Roger Penrose. 2 2 ^(2){ }^{2}2 This involves classifying different types of infinity according to whether they represent the limiting behaviour of null, timelike, or spacelike curves. In order to understand infinities, we use coordinate transformations to map infinities back to finite coordinates. We then compare the structure of the resulting metrics using the idea of conformal structure. 3 3 ^(3){ }^{3}3 This allows us to remove some of the extraneous geometrical baggage in our descriptions and simply analyse the topology of the universes via their conformal structure. In this spirit, the conformal structure can be presented using a Penrose diagram, which is a cartoon of the infinities in a space that presents their causal consequences. 4 4 ^(4){ }^{4}4 Although the origin of the mathematical tricks used to generate Penrose diagrams can sometimes look obscure, the idea is simple: a Penrose diagram represents all of spacetime in a two-dimensional sketch. Its boundaries are the infinities and light travels along lines 45 45 45^(@)45^{\circ}45 to the vertical, making the light-cone structure look like that of flat space. Let's make a start investigating infinities in spacetime.

19.1 Penrose diagrams

The first tool we shall need is a method of comparing model universes. We are interested in the causal structure of our models, encoded in the arrangement of light cones. We therefore seek a method to compare metrics that preserves this light-cone structure, and hence also the causal structure of the universe. This is where conformal structure comes in.
Two metrics g g g\boldsymbol{g}g and g ^ g ^ hat(g)\hat{\boldsymbol{g}}g^ are said to be conformally related if
g ^ ( x ) = Ω ( x ) 2 g ( x ) , g ^ ( x ) = Ω ( x ) 2 g ( x ) , hat(g)(x)=Omega(x)^(2)g(x),\hat{\boldsymbol{g}}(x)=\Omega(x)^{2} \boldsymbol{g}(x),g^(x)=Ω(x)2g(x),
(19.1)
for some differentiable function of spacetime coordinates Ω ( x ) Ω ( x ) Omega(x)\Omega(x)Ω(x), known as a conformal factor. Equation 19.1 allows us to write
(19.2) g ^ ( x , y ) g ^ ( v , w ) = g ( x , y ) g ( v , w ) (19.2) g ^ ( x , y ) g ^ ( v , w ) = g ( x , y ) g ( v , w ) {:(19.2)(( hat(g))(x,y))/(( hat(g))(v,w))=(g(x,y))/(g(v,w)):}\begin{equation*} \frac{\hat{\boldsymbol{g}}(\boldsymbol{x}, \boldsymbol{y})}{\hat{\boldsymbol{g}}(\boldsymbol{v}, \boldsymbol{w})}=\frac{\boldsymbol{g}(\boldsymbol{x}, \boldsymbol{y})}{\boldsymbol{g}(\boldsymbol{v}, \boldsymbol{w})} \tag{19.2} \end{equation*}(19.2)g^(x,y)g^(v,w)=g(x,y)g(v,w)
for vectors x , y , v x , y , v x,y,v\boldsymbol{x}, \boldsymbol{y}, \boldsymbol{v}x,y,v and w w w\boldsymbol{w}w. With these definitions, both (i) the angles between vectors, 5 5 ^(5){ }^{5}5 and (ii) the ratios of the magnitudes of vectors, are identical for two conformally related metrics. 6 6 ^(6){ }^{6}6 A null vector in a metric therefore remains null in a conformally related metric, so the light-cone structure of the spacetime described by both metrics is identical. We conclude that using this idea of conformally related metrics to compare the structure of spacetimes allows us to investigate the causal structure and infinities in our cosmological models.
Example 19.1
We have met hyperbolic spacetimes several times. In the hyperbolic (sometimes called Lobachevskian) geometry, Euclid's parallel postulate fails and the internal angles of triangles sum to less than π π pi\piπ. The hyperbolic plane is represented using the Poincaré disc in Fig. 19.3, a mathematical image inspired by the famous woodcut by M. C. Escher (1898-1972). Notice how the straight lines of this geometry all meet the circular boundary at right angles, and each black triangle is in an equivalent environment to every other. The representation shown in this figure is a conformal one, which ensures that all of the angles in the triangles are maintained throughout the picture.
Before ploughing into the details of conformal transformations, let's start to think intuitively about spacetime, and begin with flat Minkowski spacetime. This, as you will recall, involves three space coordinates, which we will simplify by considering a single radial coordinate r r rrr, and one time coordinate t t ttt, allowing us to represent Minkowski space on a two-dimensional plane (strictly a half-plane, as the radial coordinate r > 0 r > 0 r > 0r>0r>0 ), as shown in Fig. 19.4. In such a diagram, the horizontal lines represent spacelike surfaces; vertical lines represent timelike surfaces; diagonal lines lying at 45 45 45^(@)45^{\circ}45 to these directions (shown as dotted lines in Fig. 19.4) are null surfaces, on which photons can travel.
Buzz Lightyear may have said 'To Infinity, and beyond!', but where exactly is infinity? Standing at the origin in our Minkowski diagram, we can identify five distinct types of infinity. These are shown in Fig. 19.5(a), and we will go through them carefully in turn.
  • Standing at the origin, we can think about the far, far future. What will happen as t t ttt goes to infinity at our present location? This is a future we can potentially influence, as it falls within our light cone. We will label this point in our infinite future by the symbol i + i + i^(+)i^{+}i+and call it the future timelike infinity.
5 The cosine angle between vectors x x x\boldsymbol{x}x
and y y y\boldsymbol{y}y is defined in terms of metric as
(19.3) g ( x , y ) | g ( x , x ) g ( y , y ) | 1 2 . (19.3) g ( x , y ) | g ( x , x ) g ( y , y ) | 1 2 {:(19.3)(g(x,y))/(|g(x,x)g(y,y)|^((1)/(2)))". ":}\begin{equation*} \frac{g(x, y)}{|g(x, x) g(y, y)|^{\frac{1}{2}}} \text {. } \tag{19.3} \end{equation*}(19.3)g(x,y)|g(x,x)g(y,y)|12
6 6 ^(6){ }^{6}6 See exercises,
Fig. 19.3 A conformal representation of the hyperbolic plane.
Fig. 19.4 A simple representation of Minkowski spacetime with a radial coordinate r r rrr and a time coordinate t t ttt. Horizontal lines are spacelike surfaces. Vertical lines are timelike surfaces. The diagonal lines (dotted) are null surfaces, on which photons can travel.
7 7 ^(7){ }^{7}7 The symbol I + I + I^(+)\mathscr{I}^{+}I+is a script I and is pronounced 'scri plus', when said out loud.
8 8 ^(8){ }^{8}8 In other words, we need to find a conformal mapping of the Minkowsk spacetime coordinates ( t , r ) ( t , r ) (t,r)(t, r)(t,r) into some new coordinates ( t , r ) t , r (t^('),r^('))\left(t^{\prime}, r^{\prime}\right)(t,r) that sends Fig. 19.5(a) into Fig. 19.5(b). We will work out the mathematical details in a couple of pages time, but for now we will simply assume that it can be done.
Fig. 19.5 (a) The different types of infinity in Minkowski spacetime. (b) The same information, now plotted on a Penrose diagram. All the points in Minkowski space are mapped into the shaded region in the Penrose diagram.
  • Similarly, we can ask where we came from in the far, far past. In flat Minkowski spacetime, there is no Big Bang. The Universe was always here and so out there in the distant past, shown by the downward arrows in Fig. 19.5(a), is a point we will label i i i^(-)i^{-}i, the past timelike infinity.
  • We can also identify a region far away from us in a spatial sense, out at infinite distance. Thus, our arrows point somewhere over to the right. This region has no causal connection to us, since light would never reach us from this distance, and so we will label this point i 0 i 0 i^(0)i^{0}i0 and call it spacelike infinity.
  • What if we send light rays out from around the origin? They will travel on lines at 45 45 45^(@)45^{\circ}45 to the horizontal and will end up in the region indicated by the arrows travelling towards a surface labelled 7 I + 7 I + ^(7)I^(+){ }^{7} \mathscr{I}^{+}7I+, known as the future null infinity.
  • We can also receive light signals from a surface I I I^(-)\mathscr{I}^{-}Icalled the past null infinity.
Pointing in the general direction of these five different infinities is not a very grown-up way of dealing with the mathematics. What we want to do is to find a way of mapping infinite Minkowski space into a finite region, 8 8 ^(8){ }^{8}8 and this will result in the Penrose diagram shown in Fig. 19.5(b). The Penrose diagram [Fig. 19.5(b)] has exactly the same information as the Minkowski space picture [Fig. 19.5(a)], but the points at infinity have been mapped to particular points ( i + , i 0 i + , i 0 (i^(+),i^(0):}\left(i^{+}, i^{0}\right.(i+,i0 and i ) i {:i^(-))\left.i^{-}\right)i)or lines ( I + I + I^(+)\mathscr{I}^{+}I+ and I I I^(-)\mathscr{I}^{-}I) in the Penrose diagram. The whole of Minkowski space has
been squeezed into the shaded region in Fig. 19.5(b), which means that the horizontal and vertical lines in Minkowski space are now mapped into curves in the Penrose diagram. The boundary of the shaded region represents the 'dangerous' infinities of the Minkowski spacetime, which are now 'tamed' (well, at least made finite) in the Penrose diagram.
(a)

(b)
Fig. 19.7 (a) The Penrose diagram for Minkowski space is criss-crossed by timelike and spacelike surfaces. (b) The null surfaces, showing how rays of light travel, remain straight lines and connect I I I^(-)\mathscr{I}^{-}Iand I + I + I^(+)\mathscr{I}^{+}I+.
Let's see how particular features in Minkowski space are mapped into the Penrose diagram.
  • Vertical lines in Minkowski space (the timelike surfaces) are mapped to the curves shown in Fig. 19.6(a). Each curve starts at i i i^(-)i^{-}iand ends at i + i + i^(+)i^{+}i+. Since all massive particles fall along timelike geodesics, we note that every particle's world line must also start at i i i^(-)i^{-}iand end at i + i + i^(+)i^{+}i+.
  • Horizontal lines in Minkowski space (the spacelike surfaces) are mapped to the curves shown in Fig. 19.6(a). All spacelike geodesics start and end at i 0 i 0 i^(0)i^{0}i0 (remember that r r rrr is a radial coordinate, and so the spacelike geodesic starts at i 0 i 0 i^(0)i^{0}i0, goes through the origin and heads back out to i 0 i 0 i^(0)i^{0}i0 ).
  • Collecting these two results, the horizontal and vertical lines of Minkowski space (Fig. 19.4) become the curves of Fig. 19.7(a).
  • Light rays from any point in Minkowski space are lines at ± 45 ± 45 +-45^(@)\pm 45^{\circ}±45, and our mapping will preserve this feature in the Penrose diagram. Consider first the forward light cone of a point on the vertical axis ( r = 0 ) r = 0 (r^(')=0)\left(r^{\prime}=0\right)(r=0). Outgoing, future-directed rays meet the surface I + I + I+\mathscr{I}+I+ as shown in Fig 19.7(b). Incoming, future-directed rays meet the vertical axis. However, since the diagram is symmetric about
Fig. 19.6 The Penrose diagram for Minkowski space showing (a) timelike surfaces and (b) spacelike surfaces. (Surfaces shown are for r r r^(')r^{\prime}r and t = t = t^(')=t^{\prime}=t= 0.5 , 1 , 1.5 , 2 0.5 , 1 , 1.5 , 2 0.5,1,1.5,20.5,1,1.5,20.5,1,1.5,2 and 2.5.)
Fig. 19.8 The Penrose diagram for Minkowski space, showing the backward light cone from point P P P\mathcal{P}P.
Fig. 19.9 Minkowski spacetime in light-cone coordinates. Each point represents a sphere of radius r r rrr.
the vertical axis when considered in three dimensions, the forward light cone from any point meets some part of the boundary at I + I + I^(+)\mathscr{I}^{+}I+. Similarly, all rays forming the backward light cone meet the surface I I I^(-)\mathscr{I}^{-}Iand, therefore, all light rays must start at past null infinity I I I^(-)\mathscr{I}^{-}Iand end up at future null infinity I + I + I^(+)\mathscr{I}^{+}I+. Thus, we conclude that future-directed null geodesics all start on I I I^(-)\mathscr{I}^{-}Iand end on I + I + I^(+)\mathscr{I}^{+}I+.
Example 19.2
Now consider the state of affairs shown in Fig. 19.8. The shaded region shows the parts of spacetime from which light signals can reach point P P P\mathcal{P}P. An observer at P P P\mathcal{P}P can only have knowledge of events that have occurred in the shaded region, but has no knowledge of events outside this region. Since all timelike geodesics must end at i + i + i^(+)i^{+}i+, this implies that in Minkowski space an observer on a timelike geodesics eventually reaches a point where light signals can have reached them from any point in spacetime. It is also the case that the backward light cone from P P P\mathcal{P}P always includes in spacetime. It is also the case that the backward light cone from P P P\mathcal{P}P always includes
i i i^(-)i^{-}iand so there are no particles in the Minkowski Universe from whom the observer i i i^(-)i^{-}iand so there are no particles in the Minkowski Universe from whom the observer
at P P P\mathcal{P}P can't receive light signals if she waits long enough. The observer in Minkowski space can therefore potentially see every particle in the universe.
Now let's try and work out some of the details of the conformal mapping. As a first step, we express Minkowski space in light-cone coordinates.

Example 19.3

Minkowski space has the usual metric η η eta\boldsymbol{\eta}η with line element
(19.4) d s 2 = d t 2 + d x 2 + d y 2 + d z 2 (19.4) d s 2 = d t 2 + d x 2 + d y 2 + d z 2 {:(19.4)ds^(2)=-dt^(2)+dx^(2)+dy^(2)+dz^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2} \tag{19.4} \end{equation*}(19.4)ds2=dt2+dx2+dy2+dz2
In spherical polar coordinates, this becomes
(19.5) d s 2 = d t 2 + d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (19.5) d s 2 = d t 2 + d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(19.5)ds^(2)=-dt^(2)+dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.5} \end{equation*}(19.5)ds2=dt2+dr2+r2( dθ2+sin2θ dϕ2)
Since we're interested in the light-cone structure, we re-express the metric line element in light-cone coordinates
(19.6) u = t r , v = t + r (19.6) u = t r , v = t + r {:(19.6)u=t-r","quad v=t+r:}\begin{equation*} u=t-r, \quad v=t+r \tag{19.6} \end{equation*}(19.6)u=tr,v=t+r
where r 2 = x 2 + y 2 + z 2 r 2 = x 2 + y 2 + z 2 r^(2)=x^(2)+y^(2)+z^(2)r^{2}=x^{2}+y^{2}+z^{2}r2=x2+y2+z2. In terms of these coordinates, outgoing light rays (which have r r rrr increasing and t t ttt increasing) are lines of constant u u uuu, while incoming light rays ( r r rrr decreasing for increasing time t t ttt ) are lines of constant v v vvv. If we express the Minkowski metric in light-cone coordinates, we obtain
(19.7) d s 2 = d u d v + 1 4 ( v u ) 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (19.7) d s 2 = d u d v + 1 4 ( v u ) 2 d θ 2 + sin 2 θ d ϕ 2 {:(19.7)ds^(2)=-dudv+(1)/(4)(v-u)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} u \mathrm{~d} v+\frac{1}{4}(v-u)^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.7} \end{equation*}(19.7)ds2=du dv+14(vu)2( dθ2+sin2θ dϕ2)
The part of the metric ( d θ 2 + sin 2 θ d ϕ 2 ) d θ 2 + sin 2 θ d ϕ 2 (dtheta^(2)+sin^(2)theta(d)phi^(2))\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)(dθ2+sin2θ dϕ2) is simply that of a 2 -sphere. If we suppress two dimensions, as we did when discussing Minkowski spacetime, then we can draw the two-dimensional plot of this spacetime shown in Fig. 19.9, as long as we interpret each point as representing a 2 -sphere of radius r r rrr.
Fig. 19.10 The function y = tan 1 x y = tan 1 x y=tan^(-1)xy=\tan ^{-1} xy=tan1x can be used to map infinities in x x xxx back to an angle-like variable y y yyy.

Example 19.4

To examine an infinity in the Minkowski metric η η eta\boldsymbol{\eta}η, we could try the simple tactic of using a new coordinate V = 1 / v V = 1 / v V=1//vV=1 / vV=1/v, since this maps infinite v v vvv to V = 0 V = 0 V=0V=0V=0. However, if we try substituting this we end up with a line element for Minkowski space of
(19.8) d s 2 = 1 V 2 d u d V + 1 4 ( 1 V u ) 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (19.8) d s 2 = 1 V 2 d u d V + 1 4 1 V u 2 d θ 2 + sin 2 θ d ϕ 2 {:(19.8)ds^(2)=(1)/(V^(2))*dudV+(1)/(4)((1)/(V)-u)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=\frac{1}{V^{2}} \cdot \mathrm{~d} u \mathrm{~d} V+\frac{1}{4}\left(\frac{1}{V}-u\right)^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.8} \end{equation*}(19.8)ds2=1V2 du dV+14(1Vu)2( dθ2+sin2θ dϕ2)
This metric is still the Minkowski metric η η eta\boldsymbol{\eta}η, but in these coordinates it is singular at V = 0 V = 0 V=0V=0V=0, where the line element blows up. This first attempt has failed.
Undeterred, we make a sidestep and consider a new, unphysical metric g g ¯ bar(g)\overline{\boldsymbol{g}}g such that
(19.9) η = 1 V 2 g (19.9) η = 1 V 2 g ¯ {:(19.9)eta=(1)/(V^(2)) bar(g):}\begin{equation*} \boldsymbol{\eta}=\frac{1}{V^{2}} \overline{\boldsymbol{g}} \tag{19.9} \end{equation*}(19.9)η=1V2g
which is to say that the two metrics η η eta\boldsymbol{\eta}η and g g ¯ bar(g)\overline{\boldsymbol{g}}g are conformally related (i.e. g = Ω 2 η g ¯ = Ω 2 η bar(g)=Omega^(2)eta\overline{\boldsymbol{g}}=\Omega^{2} \boldsymbol{\eta}g=Ω2η with a conformal factor Ω ( x ) = V Ω ( x ) = V Omega(x)=V\Omega(x)=VΩ(x)=V ). As a result, the line element of the unphysical metric is given by d s ¯ 2 = V 2 d s 2 d s ¯ 2 = V 2 d s 2 d bar(s)^(2)=V^(2)ds^(2)\mathrm{d} \bar{s}^{2}=V^{2} \mathrm{~d} s^{2}ds¯2=V2 ds2 or
(19.10) d s ¯ 2 = d u d V + 1 4 ( 1 u V ) 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (19.10) d s ¯ 2 = d u d V + 1 4 ( 1 u V ) 2 d θ 2 + sin 2 θ d ϕ 2 {:(19.10)d bar(s)^(2)=dudV+(1)/(4)(1-uV)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} \bar{s}^{2}=\mathrm{d} u \mathrm{~d} V+\frac{1}{4}(1-u V)^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.10} \end{equation*}(19.10)ds¯2=du dV+14(1uV)2( dθ2+sin2θ dϕ2)
which is well behaved at V = 0 . 9 V = 0 . 9 V=0.^(9)V=0 .{ }^{9}V=0.9 Since the unphysical metric g g ¯ bar(g)\overline{\boldsymbol{g}}g and the Minkowski metric η η eta\boldsymbol{\eta}η are related by a conformal transformation, they share the same light-cone structure and so we can equally well study g g ¯ bar(g)\overline{\boldsymbol{g}}g as a more convenient alternative to η η eta\boldsymbol{\eta}η.
Our strategy is now clear: (i) use coordinates that map infinities to finite points and then (ii) identify an unphysical, conformally related metric that does not have any problems with its coordinates. We can then study the unphysical metric since the conformal relationship guarantees it has the same causal structure.
Example 19.5
The same strategy employed in the last example can be enacted more cleverly by defining g = Ω 2 η g ¯ = Ω 2 η bar(g)=Omega^(2)eta\overline{\boldsymbol{g}}=\Omega^{2} \boldsymbol{\eta}g=Ω2η with
(19.11) Ω 2 = 4 ( 1 + v 2 ) ( 1 + u 2 ) (19.12) v = tan p , u = tan q (19.11) Ω 2 = 4 1 + v 2 1 + u 2 (19.12) v = tan p , u = tan q {:[(19.11)Omega^(2)=(4)/((1+v^(2))(1+u^(2)))],[(19.12)v=tan p","quad u=tan q]:}\begin{align*} \Omega^{2} & =\frac{4}{\left(1+v^{2}\right)\left(1+u^{2}\right)} \tag{19.11}\\ v & =\tan p, \quad u=\tan q \tag{19.12} \end{align*}(19.11)Ω2=4(1+v2)(1+u2)(19.12)v=tanp,u=tanq
Now define p p ppp and q q qqq by
Fig. 19.11 The Einstein static Universe represented on a cylinder to demonstrate its topology. Each circle represents a 3 -sphere.
10 10 ^(10){ }^{10}10 This is achieved technically by embedding the cylinder x 2 + y 2 = 1 x 2 + y 2 = 1 x^(2)+y^(2)=1x^{2}+y^{2}=1x2+y2=1 in (2+1)-dimensional Minkowski spacetime. The full spacetime can be embed ded as the cylinder x 2 + y 2 + z 2 + w 2 = 1 x 2 + y 2 + z 2 + w 2 = 1 x^(2)+y^(2)+z^(2)+w^(2)=1x^{2}+y^{2}+z^{2}+w^{2}=1x2+y2+z2+w2=1 in (4+1)-dimensional Minkowski spacetime (see Appendix D).
Fig. 19.12 The causal structure of Minkowski space mapped onto a limited region of a larger space. The larger space is the Einstein static Universe.
where
(19.13) π / 2 < p < π / 2 , π / 2 < q < π / 2 , p q (19.13) π / 2 < p < π / 2 , π / 2 < q < π / 2 , p q {:(19.13)-pi//2 < p < pi//2","quad-pi//2 < q < pi//2","quad p >= q:}\begin{equation*} -\pi / 2<p<\pi / 2, \quad-\pi / 2<q<\pi / 2, \quad p \geq q \tag{19.13} \end{equation*}(19.13)π/2<p<π/2,π/2<q<π/2,pq
Using sec 2 θ = 1 + tan 2 θ sec 2 θ = 1 + tan 2 θ sec^(2)theta=1+tan^(2)theta\sec ^{2} \theta=1+\tan ^{2} \thetasec2θ=1+tan2θ, the function Ω 2 Ω 2 Omega^(2)\Omega^{2}Ω2 becomes Ω 2 = 4 / ( sec 2 p sec 2 q ) Ω 2 = 4 / sec 2 p sec 2 q Omega^(2)=4//(sec^(2)psec^(2)q)\Omega^{2}=4 /\left(\sec ^{2} p \sec ^{2} q\right)Ω2=4/(sec2psec2q), and the metric can be rewritten as
d s 2 = sec 2 p sec 2 q [ d p d q + 1 4 sin 2 ( p q ) ( d θ 2 + sin 2 θ d ϕ 2 ) ] (19.14) = 1 Ω 2 [ 4 d p d q + sin 2 ( p q ) ( d θ 2 + sin 2 θ d ϕ 2 ) ] . d s 2 = sec 2 p sec 2 q d p d q + 1 4 sin 2 ( p q ) d θ 2 + sin 2 θ d ϕ 2 (19.14) = 1 Ω 2 4 d p d q + sin 2 ( p q ) d θ 2 + sin 2 θ d ϕ 2 . {:[ds^(2)=sec^(2)psec^(2)q[-dp(d)q+(1)/(4)sin^(2)(p-q)(dtheta^(2)+sin^(2)theta(d)phi^(2))]],[(19.14)=(1)/(Omega^(2))[-4(d)p(d)q+sin^(2)(p-q)(dtheta^(2)+sin^(2)theta(d)phi^(2))].]:}\begin{align*} \mathrm{d} s^{2} & =\sec ^{2} p \sec ^{2} q\left[-\mathrm{d} p \mathrm{~d} q+\frac{1}{4} \sin ^{2}(p-q)\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \\ & =\frac{1}{\Omega^{2}}\left[-4 \mathrm{~d} p \mathrm{~d} q+\sin ^{2}(p-q)\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] . \tag{19.14} \end{align*}ds2=sec2psec2q[dp dq+14sin2(pq)(dθ2+sin2θ dϕ2)](19.14)=1Ω2[4 dp dq+sin2(pq)(dθ2+sin2θ dϕ2)].
We therefore identify the line element d s ¯ d s ¯ d bar(s)\mathrm{d} \bar{s}ds¯ of the unphysical metric g g ¯ bar(g)\overline{\boldsymbol{g}}g as being given by
(19.15) d s ¯ 2 = 4 d p d q + sin 2 ( p q ) ( d θ 2 + sin 2 θ d ϕ 2 ) (19.15) d s ¯ 2 = 4 d p d q + sin 2 ( p q ) d θ 2 + sin 2 θ d ϕ 2 {:(19.15)d bar(s)^(2)=-4dpdq+sin^(2)(p-q)(dtheta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} \bar{s}^{2}=-4 \mathrm{~d} p \mathrm{~d} q+\sin ^{2}(p-q)\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.15} \end{equation*}(19.15)ds¯2=4 dp dq+sin2(pq)(dθ2+sin2θ dϕ2)
This is the metric whose causal structure can be investigated. The line element d s ¯ 2 d s ¯ 2 d bar(s)^(2)\mathrm{d} \bar{s}^{2}ds¯2 can be reduced to a more familiar form by defining
(19.16) t = p + q , r = p q (19.16) t = p + q , r = p q {:(19.16)t^(')=p+q","quadr^(')=p-q:}\begin{equation*} t^{\prime}=p+q, \quad r^{\prime}=p-q \tag{19.16} \end{equation*}(19.16)t=p+q,r=pq
where
(19.17) π < t + r < π , π < t r < π , r 0 (19.17) π < t + r < π , π < t r < π , r 0 {:(19.17)-pi < t^(')+r^(') < pi","quad-pi < t^(')-r^(') < pi","quadr^(') >= 0:}\begin{equation*} -\pi<t^{\prime}+r^{\prime}<\pi, \quad-\pi<t^{\prime}-r^{\prime}<\pi, \quad r^{\prime} \geq 0 \tag{19.17} \end{equation*}(19.17)π<t+r<π,π<tr<π,r0
With this choice of coordinates, the metric becomes
(19.18) d s ¯ 2 = ( d t ) 2 + ( d r ) 2 + sin 2 r ( d θ 2 + sin 2 θ d ϕ 2 ) (19.18) d s ¯ 2 = d t 2 + d r 2 + sin 2 r d θ 2 + sin 2 θ d ϕ 2 {:(19.18)d bar(s)^(2)=-(dt^('))^(2)+(dr^('))^(2)+sin^(2)r^(')(dtheta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} \bar{s}^{2}=-\left(\mathrm{d} t^{\prime}\right)^{2}+\left(\mathrm{d} r^{\prime}\right)^{2}+\sin ^{2} r^{\prime}\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.18} \end{equation*}(19.18)ds¯2=(dt)2+(dr)2+sin2r(dθ2+sin2θ dϕ2)
This is the line element of Universe 8 (the Einstein static Universe, a closed k = 1 k = 1 k=1k=1k=1 cosmology) which has a metric line element
(19.19) d s 2 = d t 2 + d χ 2 + sin 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) (19.19) d s 2 = d t 2 + d χ 2 + sin 2 χ d θ 2 + sin 2 θ d ϕ 2 {:(19.19)ds^(2)=-dt^(2)+dchi^(2)+sin^(2)chi((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{d} \chi^{2}+\sin ^{2} \chi\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.19} \end{equation*}(19.19)ds2=dt2+dχ2+sin2χ( dθ2+sin2θ dϕ2)
and has spatial hypersurfaces that are all spherical. Specifically, these are 3 -spheres covered by the coordinates
(19.20) 0 χ π , 0 θ π , π < ϕ π . (19.20) 0 χ π , 0 θ π , π < ϕ π . {:(19.20)0 <= chi <= pi","quad0 <= theta <= pi","quad-pi < phi <= pi.:}\begin{equation*} 0 \leq \chi \leq \pi, \quad 0 \leq \theta \leq \pi, \quad-\pi<\phi \leq \pi . \tag{19.20} \end{equation*}(19.20)0χπ,0θπ,π<ϕπ.
Since the Einstein model is static, the spheres are unchanging in time. Note that eqn 19.18 is defined over a different set of coordinates to eqn 19.19.
It will turn out that the Einstein universe will be a very useful model against which to compare our other model spacetimes. Let's therefore attempt to draw Einstein spacetime with its unchanging 3 -spheres. As you'll appreciate, it's a challenge to draw anything more complicated than a representation of a three-dimensional object on a two-dimensional page. Nonetheless, owing to the spherical symmetry of the model, we can suppress two of the spatial dimensions and just consider circles (or 1 -spheres, if you like) as a function of time. The overall shape or topology of the spacetime manifold now comprises circles of constant radius (representing the 3spheres) repeated as a function of time. We say that the topology of the manifold is cylindrical. The space can be represented 10 10 ^(10){ }^{10}10 in the diagram shown in Fig. 19.11, where θ θ theta\thetaθ and ϕ ϕ phi\phiϕ have been suppressed. We notice the awkward way that the χ χ chi\chiχ variable where θ θ theta\thetaθ and ϕ ϕ phi\phiϕ have been suppressed. We notice the awkward way that the χ χ chi\chiχ vat
increases in two directions, reflecting the fact that it varies between 0 and π π pi\piπ.
Although it has involved a certain amount of abstraction, in our cylindrical representation of the Einstein Universe, we have a structure with which we can now compare other spacetimes.
The last example shows how the whole of Minkowski space can be mapped to a limited region of the Einstein static metric. This allows us to draw a representation of Minkowski space on the cylinder in Fig. 19.11. The result is Fig. 19.12. The fact that it only covers a limited region means that the boundary of Minkowski space represents the infinities of this space. Strictly, it represents the conformal structure of infinity. The structure in Fig. 19.12 can be pared down to its bare essentials, and it is this that leads to the Penrose diagram that we saw in Fig. 19.5(b) which depicts the boundary in Fig. 19.12 plotted in the ( t , r ) t , r (t^('),r^('))\left(t^{\prime}, r^{\prime}\right)(t,r) plane, not including the superfluous half where r r r^(')r^{\prime}r increases anticlockwise around the cylinder. This approach also allows us to identify some different types of infinity that we encountered earlier.
Example 19.6
Referring to the labelled points in Fig. 19.12 we see that, in terms of the variables ( t , r ) t , r (t^('),r^('))\left(t^{\prime}, r^{\prime}\right)(t,r), we have labelled
(19.21) i 0 = ( 0 , π ) , i + = ( π , 0 ) , i = ( π , 0 ) (19.21) i 0 = ( 0 , π ) , i + = ( π , 0 ) , i = ( π , 0 ) {:(19.21)i^(0)=(0","pi)","quadi^(+)=(pi","0)","quadi^(-)=(-pi","0):}\begin{equation*} i^{0}=(0, \pi), \quad i^{+}=(\pi, 0), \quad i^{-}=(-\pi, 0) \tag{19.21} \end{equation*}(19.21)i0=(0,π),i+=(π,0),i=(π,0)
It follows that the line labelled I + I + I^(+)\mathscr{I}^{+}I+is the line t + r = π t + r = π t+r=pit+r=\pit+r=π and I I I^(-)\mathscr{I}^{-}Iis the line t r = π t r = π t-r=-pit-r=-\pitr=π. These lines, shown in Fig. 19.12 (which actually represents surfaces in spacetime, owing to our having suppressed two dimensions) are null. Since future-directed light rays have t r = 2 q = t r = 2 q = t^(')-r^(')=2q=t^{\prime}-r^{\prime}=2 q=tr=2q= const., then all future-directed light rays terminate on I + I + I^(+)\mathscr{I}^{+}I+. Similarly, the line I I I^(-)\mathscr{I}^{-}Ialso represents a null surface and all inward or past-directed light cones terminate on it (i.e. rays for which t + r = 2 p = t + r = 2 p = t^(')+r^(')=2p=t^{\prime}+r^{\prime}=2 p=t+r=2p= const).
The definitions and rules introduced so far are applied in the next section. Suitably expanded and generalized, can be used as the foundation of a set of global techniques that can also be used to make very powerful and general arguments about singularities. 11 11 ^(11){ }^{11}11

19.2 The de Sitter spacetime

Now that we have a picture of Minkowski space and its conformal structure, let's examine Universe 1 (de Sitter spacetime). This is the spacetime with flat and expanding space described by line element
(19.22) d s = d t 2 + e 2 H t [ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (19.22) d s = d t 2 + e 2 H t d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(19.22)ds=-dt^(2)+e^(2Ht)[(d)r^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s=-\mathrm{d} t^{2}+\mathrm{e}^{2 H t}\left[\mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{19.22} \end{equation*}(19.22)ds=dt2+e2Ht[ dr2+r2( dθ2+sin2θ dϕ2)]
With a little algebra it can be shown 12 12 ^(12){ }^{12}12 that the line element for Universe 1 can be rewritten in terms of a new timelike variable t t t^(')t^{\prime}t as
(19.23) d s 2 = α 2 sec 2 t d s ¯ 2 , (19.23) d s 2 = α 2 sec 2 t d s ¯ 2 , {:(19.23)ds^(2)=alpha^(2)sec^(2)t^(')d bar(s)^(2)",":}\begin{equation*} \mathrm{d} s^{2}=\alpha^{2} \sec ^{2} t^{\prime} \mathrm{d} \bar{s}^{2}, \tag{19.23} \end{equation*}(19.23)ds2=α2sec2tds¯2,
where d s ¯ 2 d s ¯ 2 d bar(s)^(2)\mathrm{d} \bar{s}^{2}ds¯2 is the line element for the Einstein Universe [with coordinates ( t , r , θ , ϕ ) ] t , r , θ , ϕ {:(t^('),r,theta,phi)]\left.\left(t^{\prime}, r, \theta, \phi\right)\right](t,r,θ,ϕ)], where π 2 < t < π 2 , α 2 = 1 / H 2 π 2 < t < π 2 , α 2 = 1 / H 2 -(pi)/(2) < t^(') < (pi)/(2),alpha^(2)=1//H^(2)-\frac{\pi}{2}<t^{\prime}<\frac{\pi}{2}, \alpha^{2}=1 / H^{2}π2<t<π2,α2=1/H2, and the other coordinates take the same values as in the Einstein static Universe. This means that de Sitter space is conformal to the Einstein static Universe, but only for times π 2 < t < π 2 π 2 < t < π 2 -(pi)/(2) < t^(') < (pi)/(2)-\frac{\pi}{2}<t^{\prime}<\frac{\pi}{2}π2<t<π2.
The resulting Penrose diagram is shown in Fig. 19.13. Strikingly, there are no unique timelike ( i ± ) i ± (i^(+-))\left(i^{ \pm}\right)(i±)or spacelike ( i 0 ) i 0 (i^(0))\left(i^{0}\right)(i0) infinities on which world lines terminate, as we had for Minkowski spacetime. Instead, there is a past null infinity surface I I I^(-)\mathscr{I}^{-}Iand a future null infinity surface I + I + I^(+)\mathscr{I}^{+}I+. These are spacelike: they link up all of χ χ chi\chiχ at a single instant ( t = ± π / 2 ) t = ± π / 2 (t^(')=+-pi//2)\left(t^{\prime}= \pm \pi / 2\right)(t=±π/2). As a result of this difference in conformal structure, causality in de Sitter spacetime is rather different to that in Minkowski spacetime and provides the possibility of some new features: particle horizons and event horizons. Since, in de Sitter spacetime, all timelike geodesics start at I I I^(-)\mathscr{I}^{-}Iand end at I + I + I^(+)\mathscr{I}^{+}I+, the past light cone of an observer at a point p p ppp does not, generally, capture all points in the past. This means that there are events, and even whole world lines of events, that an observer never sees. An example is shown in Fig. 19.14: some free particles fall
11 11 ^(11){ }^{11}11 We discuss an example in Chapter 50. See Penrose (1973) for the full story.
12 12 ^(12){ }^{12}12 See exercises.
Fig. 19.13 Penrose diagram for de Sitter spacetime.
Fig. 19.15 The observer whose world line coincides with the origin of the radial coordinate χ = 0 χ = 0 chi=0\chi=0χ=0 at t = π / 2 t = π / 2 t^(')=pi//2t^{\prime}=\pi / 2t=π/2 has a light cone that sweeps out the whole χ χ chi\chiχ axis.
13 13 ^(13){ }^{13}13 It is perhaps tempting to think that an observer at χ = π / 2 χ = π / 2 chi=pi//2\chi=\pi / 2χ=π/2 can have seen events from the entire χ χ chi\chiχ axis at the earlier time t = 0 t = 0 t^(')=0t^{\prime}=0t=0. However, there cannot be a difference since each value of χ χ chi\chiχ is equivalent. The problem here is that χ χ chi\chiχ is a radius-like variable so it only makes sense to consider the light cone over increasing or decreasing χ χ chi\chiχ when considering the question of the furthest a light ray can travel in a given time.
Fig. 19.16 The future event horizons for an observer travelling along a world line O . The shaded region lie's outside O's future event horizon and events in this region will never be seen by 0 .
Fig. 19.14 The particle horizon for an observer at point P P P\mathcal{P}P, in a spacetime with a spacelike past null infinity I I I^(-)\mathscr{I}^{-}I. There exist particles whose world lines do not intersect the past null cone of P P P\mathcal{P}P and so are not yet visible to the observer.
along world lines that lie outside the backward light cone of the observer at P P P\mathcal{P}P. This leads to a division between those geodesics that can be seen by the observer at P P P\mathcal{P}P and those that can't. This division is known as the particle horizon at P P P\mathcal{P}P. As the observer moves along their world line, they are able to see more and more particle geodesics, so this horizon is observer dependent. Moreover, when the observer in de Sitter space finally meets the future null infinity I + I + I^(+)\mathscr{I}^{+}I+, their past light cone does not sweep out all of spacetime. [As shown in Fig. 19.15, the backward light cone of an observer whose world line coincides with the origin of the radial coordinate χ χ chi\chiχ does sweep over all values of χ χ chi\chiχ at future infinity ( t = π / 2 ) t = π / 2 (t^(')=-pi//2)\left(t^{\prime}=-\pi / 2\right)(t=π/2) for this spacetime, allowing the observer to see, at different times, signals from all of the events occurring in the shaded region. 13 13 ^(13){ }^{13}13 ] There are, therefore, events that can never been seen by the observer. The surface formed from the backward light cone on the future infinity I + I + I^(+)\mathscr{I}^{+}I+is known as the future event horizon. By contrast, since the observer in Minkowski space can see all of spacetime when they reach i + i + i^(+)i^{+}i+, there is no future event horizon in Minkowski spacetime.
Example 19.7
Consider the Penrose diagram in Fig. 19.16. The observer at point P P P\mathcal{P}P on world line O can see signals emitted from part of a world line Q Q QQQ originating at I I I^(-)\mathscr{I}^{-}I. In other words, since part of curve Q is inside the particle horizon at P P P\mathcal{P}P, a particle travelling along world line Q Q QQQ is sometimes visible to the observer at P P P\mathcal{P}P. Owing to the existence of future event horizons in this spacetime, there is a point R R R\mathcal{R}R, where the world line Q meets the observer's future event horizon. The observer on P P P\mathcal{P}P has no possibility of seeing events any further in the future along Q Q QQQ than R R R\mathcal{R}R. Imagine travelling along the observer's world line O O OOO. They can only see the event R R R\mathcal{R}R when they (the observer) are at infinity (since this is the first time that the event at R R R\mathcal{R}R lies on their past light cone).
Just as we have a future event horizon, we can also define a past event horizon. 14 14 ^(14){ }^{14}14 This is the surface traced out by the forward light cone of a point on the past null infinity I I I^(-)\mathscr{I}^{-}Ias also shown in Fig. 19.17. This surface divides up all of the events that can be influenced or contacted by an observer from those that can't (with events outside the past event horizon unable to be influenced by the observer).
Example 19.8
Consider the Penrose diagram shown in Fig. 19.17. Region I contains events that can be both seen and influenced by the observer; region II contains events that can be influenced, but never seen, and region III contains events that can be seen, but never influenced.

19.3 Big-Bang singularities

Finally, we turn to the conformal structure of the universes based on the Robertson-Walker geometries. The Robertson-Walker spaces are defined a metric with line element 15 15 ^(15){ }^{15}15
(19.24) d s 2 = d t 2 + a ( t ) 2 [ d χ 2 + Σ ( χ ) 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] (19.24) d s 2 = d t 2 + a ( t ) 2 d χ 2 + Σ ( χ ) 2 d θ 2 + sin 2 θ d ϕ 2 {:(19.24)ds^(2)=-dt^(2)+a(t)^(2)[(d)chi^(2)+Sigma(chi)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left[\mathrm{~d} \chi^{2}+\Sigma(\chi)^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \tag{19.24} \end{equation*}(19.24)ds2=dt2+a(t)2[ dχ2+Σ(χ)2( dθ2+sin2θ dϕ2)]
These solutions have a new feature: the initial Big-Bang singularity when a ( t = 0 ) = 0 a ( t = 0 ) = 0 a(t=0)=0a(t=0)=0a(t=0)=0. We should therefore expect that they should cover a limited region of the cylinder covered by the Einstein static Universe, and that there must be a boundary representing the Big Bang.
To investigate the conformal structure we start by using a new conformal time coordinate τ τ tau\tauτ defined by
(19.25) d t = a ( t ) d τ . (19.25) d t = a ( t ) d τ . {:(19.25)dt=a(t)dtau.:}\begin{equation*} \mathrm{d} t=a(t) \mathrm{d} \tau . \tag{19.25} \end{equation*}(19.25)dt=a(t)dτ.
The idea is that we then have d s 2 = a ( t ) 2 d s ¯ 2 d s 2 = a ( t ) 2 d s ¯ 2 ds^(2)=a(t)^(2)d bar(s)^(2)\mathrm{d} s^{2}=a(t)^{2} \mathrm{~d} \bar{s}^{2}ds2=a(t)2 ds¯2, with
(19.26) d s ¯ 2 = d τ 2 + d χ 2 + Σ ( χ ) 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (19.26) d s ¯ 2 = d τ 2 + d χ 2 + Σ ( χ ) 2 d θ 2 + sin 2 θ d ϕ 2 {:(19.26)d bar(s)^(2)=-dtau^(2)+dchi^(2)+Sigma(chi)^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} \bar{s}^{2}=-\mathrm{d} \tau^{2}+\mathrm{d} \chi^{2}+\Sigma(\chi)^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{19.26} \end{equation*}(19.26)ds¯2=dτ2+dχ2+Σ(χ)2( dθ2+sin2θ dϕ2)
This line-element is in a form that can be compared to the Einstein static Universe.
Example 19.9
We can show that the Robertson-Walker spaces with p = Λ = 0 p = Λ = 0 p=Lambda=0p=\Lambda=0p=Λ=0 are conformal to the Einstein static Universe. We first set χ = r χ = r chi=r^(')\chi=r^{\prime}χ=r and τ = t τ = t tau=t^(')\tau=t^{\prime}τ=t, where the primed coordinates are those from Example 19.5. For k = 1 k = 1 k=1k=1k=1 the line element is then conformal to the Einstein Universe but, since τ τ tau\tauτ is defined for 0 < τ < 0 < τ < 0 < tau < oo0<\tau<\infty0<τ<, we find that the space is mapped into the Einstein Universe for t = 0 t = 0 t^(')=0t^{\prime}=0t=0 (when τ = 0 τ = 0 tau=0\tau=0τ=0 ) to t = π t = π t^(')=pit^{\prime}=\pit=π (when τ = τ = tau=oo\tau=\inftyτ= ). For k = 0 k = 0 k=0k=0k=0 we have flat space, so, using the same coordinates and the procedure from Example 19.5, we again obtain a diamond-shaped region on the cylindrical surface, but this time with 0 < t < π 0 < t < π 0 < t < pi0<t<\pi0<t<π. For k = 1 k = 1 k=-1k=-1k=1 things are more tricky and we must make a coordinate transformation 16 16 ^(16){ }^{16}16 with the result that the metric becomes conformal to the region
(19.27) π 2 t + r π 2 , π 2 t r π 2 (19.27) π 2 t + r π 2 , π 2 t r π 2 {:(19.27)-(pi)/(2) <= t^(')+r^(') <= (pi)/(2)","quad-(pi)/(2) <= t^(')-r^(') <= (pi)/(2):}\begin{equation*} -\frac{\pi}{2} \leq t^{\prime}+r^{\prime} \leq \frac{\pi}{2}, \quad-\frac{\pi}{2} \leq t^{\prime}-r^{\prime} \leq \frac{\pi}{2} \tag{19.27} \end{equation*}(19.27)π2t+rπ2,π2trπ2
14 14 ^(14){ }^{14}14 Sit at a point at future infinity and draw a backward light cone get the future event horizon; sit at past infinity and cast the forward light cone to get the past event horizon.
Fig. 19.17 Past and future event horizons for a world line W W WWW, dividing up spacetime.
15 15 ^(15){ }^{15}15 The function Σ ( χ ) Σ ( χ ) Sigma(chi)\Sigma(\chi)Σ(χ) is given by
Σ ( χ ) = sin χ , ( k = + 1 ) Σ ( χ ) = sin χ , ( k = + 1 ) Sigma(chi)=sin chi,quad(k=+1)\Sigma(\chi)=\sin \chi, \quad(k=+1)Σ(χ)=sinχ,(k=+1),
Σ ( χ ) = χ , ( k = 0 ) Σ ( χ ) = χ , ( k = 0 ) Sigma(chi)=chi,quad(k=0)\Sigma(\chi)=\chi, \quad(k=0)Σ(χ)=χ,(k=0),
Σ ( χ ) = sinh χ , ( k = 1 ) Σ ( χ ) = sinh χ , ( k = 1 ) Sigma(chi)=sinh chi,quad(k=-1)\Sigma(\chi)=\sinh \chi, \quad(k=-1)Σ(χ)=sinhχ,(k=1).
16 16 ^(16){ }^{16}16 The transformation turns out to be t = tan 1 [ tanh 1 2 ( t + χ ) ] t = tan 1 tanh 1 2 ( t + χ ) t^(')=tan^(-1)[tanh((1)/(2))(t+chi)]t^{\prime}=\tan ^{-1}\left[\tanh \frac{1}{2}(t+\chi)\right]t=tan1[tanh12(t+χ)] + tan 1 [ tanh 1 2 ( t χ ) ] + tan 1 tanh 1 2 ( t χ ) +tan^(-1)[tanh((1)/(2))(t-chi)]+\tan ^{-1}\left[\tanh \frac{1}{2}(t-\chi)\right]+tan1[tanh12(tχ)], r = tan 1 [ tanh 1 2 ( t + χ ) ] r = tan 1 tanh 1 2 ( t + χ ) r^(')=tan^(-1)[tanh((1)/(2))(t+chi)]r^{\prime}=\tan ^{-1}\left[\tanh \frac{1}{2}(t+\chi)\right]r=tan1[tanh12(t+χ)] + tan 1 [ tanh 1 2 ( t χ ) ] + tan 1 tanh 1 2 ( t χ ) +tan^(-1)[tanh((1)/(2))(t-chi)]+\tan ^{-1}\left[\tanh \frac{1}{2}(t-\chi)\right]+tan1[tanh12(tχ)].
Fig. 19.19 Penrose diagram for the k = 0 k = 0 k=0k=0k=0 and k = 1 k = 1 k=-1k=-1k=1 Robertson-Walker spacetimes. The spacelike singularity is shown as a double line.
Fig. 19.20 Penrose diagram for the k = 1 k = 1 k=1k=1k=1 Robertson-Walker spacetime The spacelike singularities are depicted by double lines.
Fig. 19.18 Causal structure of the Robertson-Walker spacetimes.
The causal structure is shown on the Einstein cylinder in Fig. 19.18.
With the causal structure of the Robertson-Walker spaces determined, we draw their Penrose diagrams. As predicted, these feature a singularity. In terms of their conformal structure, the singularity is not represented as a point, but as a surface. Specifically, the RobertsonWalker spacetimes have a singular surface at t = 0 t = 0 t^(')=0t^{\prime}=0t=0 representing the Big Bang, from which all world lines must originate. This surface is spacelike, so we say that it represents a spacelike singularity.
For the flat k = 0 k = 0 k=0k=0k=0 and hyperbolic k = 1 k = 1 k=-1k=-1k=1 spaces, the Penrose diagram (Fig. 19.19) resembles the upper half of the Penrose diagram for Minkowski space, but starting from the Big-Bang singularity. Timelike geodesics start at any point on the surface I I I^(-)\mathscr{I}^{-}I, but all terminate at the point i + i + i^(+)i^{+}i+. There is therefore a past event horizon for an observer on a timelike geodesic, but there is not a future event horizon: when they reach i + i + i^(+)i^{+}i+the observer can see every event in the spacetime. (There is also a particle horizon in this spacetime.) Like massive particles, light rays also start on J J J^(-)\mathscr{J}^{-}J, but these terminate on J + J + J^(+)\mathscr{J}^{+}J+.
The situation is different for an spherical (closed) k = 1 k = 1 k=1k=1k=1 space: this is because the radial variable is constrained to lie in the range 0 < χ < π 0 < χ < π 0 < chi < pi0<\chi<\pi0<χ<π. In this case, the universe expands from an initial singularity (the Big Bang) and then contracts towards a second singularity (the Big Crunch). There are therefore two singular surfaces in the Penrose diagram, as shown in Fig. 19.20. All world lines start on the line I I I^(-)\mathscr{I}^{-}Iand terminate on I + I + I^(+)\mathscr{I}^{+}I+. Like the conformal structure of the de Sitter spacetime, the k = 1 k = 1 k=1k=1k=1 space allows past and future event horizons.

Example 19.10

Consider Universe 6 ( k = 1 , Λ = 0 ) 6 ( k = 1 , Λ = 0 ) 6(k=1,Lambda=0)6(k=1, \Lambda=0)6(k=1,Λ=0) from Chapter 18, with its Big Bang, expansion, contraction and Big Crunch, and causal structure depicted in Fig. 19.20. Since this universe is closed with Σ ( χ ) = sin χ Σ ( χ ) = sin χ Sigma(chi)=sin chi\Sigma(\chi)=\sin \chiΣ(χ)=sinχ, the χ χ chi\chiχ axis is periodic. We therefore identify χ = 0 χ = 0 chi=0\chi=0χ=0 and χ = π χ = π chi=pi\chi=\piχ=π in the Penrose diagram, so that light rays meeting the edges at χ = 0 χ = 0 chi=0\chi=0χ=0 and χ = π χ = π chi=pi\chi=\piχ=π wrap around the diagram. As shown in Fig. 19.21(a), this allows us (the observers at the origin) to see signals from all points on the χ χ chi\chiχ axis at time t = π / 2 t = π / 2 t^(')=pi//2t^{\prime}=\pi / 2t=π/2. If there were always a star at each point on the χ χ chi\chiχ grid, we would see all of the stars in the universe at this time. As shown in Fig. 19.22(a), this corresponds to radial light rays travelling to reach us by traversing the two halves of the unit circle that represents the angle-like radial coordinate χ χ chi\chiχ.
During the expansion of this universe, we therefore see more and more stars [Fig. 19.21(a)] until, at the time t = π / 2 t = π / 2 t^(')=pi//2t^{\prime}=\pi / 2t=π/2 when the universe has its maximum radius a ( t = π / 2 ) a t = π / 2 a(t^(')=pi//2)a\left(t^{\prime}=\pi / 2\right)a(t=π/2), we see all of the stars: half in front of us, and half behind. As the universe subsequently contracts, our backward light cone continues to get larger, and we start to see duplicates of the more distant stars, as light travelling over more than half of the circle begins to reach us. [An example of a duplicate is shown in Fig. 19.21(b) and Fig. 19.22(b)]. As the contraction of the universe continues, we come to see duplicates of more and more of the stars. We expect eventually to be able to see the back of our own heads (as they appeared shortly after the Big Bang!), when radial light rays from the base of our own world lines at t = 0 t = 0 t^(')=0t^{\prime}=0t=0 reach us after traversing the entire universe. However, the observation of these rays coincides with the Big Crunch, and so we are crushed to death at exactly this moment (see Exercise 19.8). Moreover, since nothing can travel faster than light, there is no possibility of any particle to traverse the whole of the closed universe.
We can also analyse the story in terms of the red- and blueshifts that the observer measures. Initially, all of the stars seen will be redshifted since a ( t em ) < a ( t ob ) a t em < a t ob a(t_(em)) < a(t_(ob))a\left(t_{\mathrm{em}}\right)<a\left(t_{\mathrm{ob}}\right)a(tem)<a(tob). During the collapsing period, there will times when starlight traversing the most direct route will be blueshifted, while light from the duplicate image, traversing the longer route, will be redshifted. During the final stages of the collapse, most of the light received will be blueshifted since a ( t em ) > a ( t ob ) a t em > a t ob a(t_(em)) > a(t_(ob))a\left(t_{\mathrm{em}}\right)>a\left(t_{\mathrm{ob}}\right)a(tem)>a(tob).
We have now seen how to describe the incorporation of singularities like the Big Bang into the conformal structure of spacetime. Singularities can now be understood in the same way that we understand infinities: they represent the boundaries of spacetime. They are different from infinities in that they could exist at a finite interval from an observer. (By definition this is not the case for an infinity!) Singularities are still potentially a cause for concern since, as they are regions where the laws of physics break down, you never know what will come out of a singularity. However, the toolkit offered by conformal structure allows us to tame them geometrically and Penrose diagrams give us a convenient way to visualize them. We shall encounter Penrose diagrams again in Chapter 25 when we examine black holes and where, once again, their usefulness in incorporating a singularity will be important.
Finally, we are finishing Part III on cosmology without saying too much about which universe we are actually in, and what evidence might exist for determining that. This is because this is a book on general relativity as a theory, rather than a primer on cosmology. But we will return to this important issue at the end of the book in Chapter 50.
Fig. 19.21 (a) In Universe 6, the observer at χ = 0 χ = 0 chi=0\chi=0χ=0 can see events from each point on the χ χ chi\chiχ grid at t = π / 2 t = π / 2 t^(')=pi//2t^{\prime}=\pi / 2t=π/2. This is because χ = 0 χ = 0 chi=0\chi=0χ=0. (b) An example of the two images of a (b) An example of the two imag
single star seen by the observer.
Fig. 19.22 (a) For the closed Universe 6 at the point of maximum expansion, light can reach us (at point A) from stars in the whole spherical univers stars in the whole spherical universe, with half going one way (appearing in front of us, say) and half going the other way (appearing behind us). (b) As the universe contracts, light from B can reach us via the direct route (anticlockwise, as shown) and the indirect route (clockwise), so we see two images of B. (Don't confuse the variable χ χ chi\chiχ with the angle around the unit circle: they differ by a factor of 2 .)

Chapter summary

  • The conformal structure of a model universe allows us to treat its infinities as points or surfaces at the edge of spacetime.
  • Penrose diagrams allow us to visualize the conformal structure of a spacetime, revealing particle and event horizons.
  • Singularities can be incorporated into Penrose diagrams: they form boundaries to spacetime that can be separated from an observer by a finite interval.

Exercises

(19.1) In Minkowski spacetime in (1+1) dimensions, light-cone coordinates are defined by the transformation
u = t x (19.28) v = t + x u = t x (19.28) v = t + x {:[u=t-x],[(19.28)v=t+x]:}\begin{align*} & u=t-x \\ & v=t+x \tag{19.28} \end{align*}u=tx(19.28)v=t+x
(a) Confirm that the metric line element, expressed in these coordinates, becomes
(19.29) d s 2 = d u d v (19.29) d s 2 = d u d v {:(19.29)ds^(2)=-dudv:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} u \mathrm{~d} v \tag{19.29} \end{equation*}(19.29)ds2=du dv
(b) How are proper lengths and proper times computed in these coordinates?
(19.2) Consider a particle moving at constant velocity along the x x xxx axis such that x = β t x = β t x=beta tx=\beta tx=βt.
(a) Transform to light-cone coordinates ( u = t u = t u=t-u=t-u=t x , v = t + x ) x , v = t + x ) x,v=t+x)x, v=t+x)x,v=t+x) and show that the light-cone velocity is
(19.30) d u d v = 1 β 1 + β (19.30) d u d v = 1 β 1 + β {:(19.30)(du)/((d)v)=(1-beta)/(1+beta):}\begin{equation*} \frac{\mathrm{d} u}{\mathrm{~d} v}=\frac{1-\beta}{1+\beta} \tag{19.30} \end{equation*}(19.30)du dv=1β1+β
(b) Investigate the light-cone velocities of particles with different values of β β beta\betaβ.
(19.3) The cosine angle between vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v is defined as
(19.31) g ( u , v ) [ g ( u , u ) g ( v , v ) ] 1 2 (19.31) g ( u , v ) [ g ( u , u ) g ( v , v ) ] 1 2 {:(19.31)(g(u,v))/([g(u,u)g(v,v)]^((1)/(2))):}\begin{equation*} \frac{g(\boldsymbol{u}, \boldsymbol{v})}{[g(\boldsymbol{u}, \boldsymbol{u}) g(\boldsymbol{v}, \boldsymbol{v})]^{\frac{1}{2}}} \tag{19.31} \end{equation*}(19.31)g(u,v)[g(u,u)g(v,v)]12
(a) Show that a conformal transformation g μ ν g μ ν g_(mu nu)rarrg_{\mu \nu} \rightarrowgμν f ( x ) g μ ν f ( x ) g μ ν f(x)g_(mu nu)f(x) g_{\mu \nu}f(x)gμν preserves all angles.
(b) Show that all null curves remain null under this transformation.
(19.4) Consider the de Sitter line element in Cartesian coordinates
d s 2 = d t ^ 2 + e 2 i α ( d x ^ 2 + d y ^ 2 + d z ^ 2 ) d s 2 = d t ^ 2 + e 2 i α d x ^ 2 + d y ^ 2 + d z ^ 2 ds^(2)=-d hat(t)^(2)+e^((2i)/(alpha))((d) hat(x)^(2)+d hat(y)^(2)+d hat(z)^(2))\mathrm{d} s^{2}=-\mathrm{d} \hat{t}^{2}+\mathrm{e}^{\frac{2 i}{\alpha}}\left(\mathrm{~d} \hat{x}^{2}+\mathrm{d} \hat{y}^{2}+\mathrm{d} \hat{z}^{2}\right)ds2=dt^2+e2iα( dx^2+dy^2+dz^2)
where α = H 1 α = H 1 alpha=H^(-1)\alpha=H^{-1}α=H1. In the exercises in Chapter 18 , we claimed that de Sitter spacetime can be represented on a hyperboloid embedded in fivedimensional Minkowski space, given by an equation
(19.33) v 2 + w 2 + x 2 + y 2 + z 2 = α 2 (19.33) v 2 + w 2 + x 2 + y 2 + z 2 = α 2 {:(19.33)-v^(2)+w^(2)+x^(2)+y^(2)+z^(2)=alpha^(2):}\begin{equation*} -v^{2}+w^{2}+x^{2}+y^{2}+z^{2}=\alpha^{2} \tag{19.33} \end{equation*}(19.33)v2+w2+x2+y2+z2=α2
(a) By making the transformations
(19.34) t ^ = α ln w + v α , x ^ = α x w + v , y ^ = α y w + v , z ^ = α z w + v , (19.34) t ^ = α ln w + v α , x ^ = α x w + v , y ^ = α y w + v , z ^ = α z w + v , {:(19.34){:[ hat(t)=alpha ln((w+v)/(alpha))",", hat(x)=(alpha x)/(w+v)","],[ hat(y)=(alpha y)/(w+v)",", hat(z)=(alpha z)/(w+v)","]:}:}\begin{array}{cl} \hat{t}=\alpha \ln \frac{w+v}{\alpha}, & \hat{x}=\frac{\alpha x}{w+v}, \tag{19.34}\\ \hat{y}=\frac{\alpha y}{w+v}, & \hat{z}=\frac{\alpha z}{w+v}, \end{array}(19.34)t^=αlnw+vα,x^=αxw+v,y^=αyw+v,z^=αzw+v,
show that d s 2 d s 2 ds^(2)\mathrm{d} s^{2}ds2 can be recast as the Minkowski line element
(19.35) d s 2 = d v 2 + d w 2 + d x 2 + d y 2 + d z 2 (19.35) d s 2 = d v 2 + d w 2 + d x 2 + d y 2 + d z 2 {:(19.35)ds^(2)=-dv^(2)+dw^(2)+dx^(2)+dy^(2)+dz^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} v^{2}+\mathrm{d} w^{2}+\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2} \tag{19.35} \end{equation*}(19.35)ds2=dv2+dw2+dx2+dy2+dz2
(b) Now consider a set of coordinates on the hyperboloid given by
v = α sinh ( H t ) w = α cosh ( H t ) cos χ (19.36) x = α cosh ( H t ) sin χ cos θ y = α cosh ( H t ) sin χ sin θ cos ϕ z = α cosh ( H t ) sin χ sin θ sin ϕ v = α sinh ( H t ) w = α cosh ( H t ) cos χ (19.36) x = α cosh ( H t ) sin χ cos θ y = α cosh ( H t ) sin χ sin θ cos ϕ z = α cosh ( H t ) sin χ sin θ sin ϕ {:[v=alpha sinh(Ht)],[w=alpha cosh(Ht)cos chi],[(19.36)x=alpha cosh(Ht)sin chi cos theta],[y=alpha cosh(Ht)sin chi sin theta cos phi],[z=alpha cosh(Ht)sin chi sin theta sin phi]:}\begin{align*} & v=\alpha \sinh (H t) \\ & w=\alpha \cosh (H t) \cos \chi \\ & x=\alpha \cosh (H t) \sin \chi \cos \theta \tag{19.36}\\ & y=\alpha \cosh (H t) \sin \chi \sin \theta \cos \phi \\ & z=\alpha \cosh (H t) \sin \chi \sin \theta \sin \phi \end{align*}v=αsinh(Ht)w=αcosh(Ht)cosχ(19.36)x=αcosh(Ht)sinχcosθy=αcosh(Ht)sinχsinθcosϕz=αcosh(Ht)sinχsinθsinϕ
Show that the de Sitter Universe takes on the metric line element
d s 2 = d t 2 + α 2 cosh 2 ( α 1 t ) × [ d χ 2 + sin 2 ϕ ( d θ 2 + sin 2 θ d ϕ 2 ) ] d s 2 = d t 2 + α 2 cosh 2 α 1 t × d χ 2 + sin 2 ϕ d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-dt^(2)+alpha^(2)cosh^(2)(alpha^(-1)t)],[ xx[dchi^(2)+sin^(2)phi((d)theta^(2)+sin^(2)theta(d)phi^(2))]]:}\begin{aligned} \mathrm{d} s^{2}= & -\mathrm{d} t^{2}+\alpha^{2} \cosh ^{2}\left(\alpha^{-1} t\right) \\ & \times\left[\mathrm{d} \chi^{2}+\sin ^{2} \phi\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)\right] \end{aligned}ds2=dt2+α2cosh2(α1t)×[dχ2+sin2ϕ( dθ2+sin2θ dϕ2)]
(c) Using the result of the previous problem, use the transformation
(19.38) tan ( t 2 + π 4 ) = e t / α (19.38) tan t 2 + π 4 = e t / α {:(19.38)tan((t^('))/(2)+(pi)/(4))=e^(t//alpha):}\begin{equation*} \tan \left(\frac{t^{\prime}}{2}+\frac{\pi}{4}\right)=\mathrm{e}^{t / \alpha} \tag{19.38} \end{equation*}(19.38)tan(t2+π4)=et/α
where π / 2 < t < π / 2 π / 2 < t < π / 2 -pi//2 < t^(') < pi//2-\pi / 2<t^{\prime}<\pi / 2π/2<t<π/2, to show that de Sitter spacetime can be recast as in eqn 19.23.
(19.5) The action for the electromagnetic field is invariant with respect to conformal transformations. Consider a Robertson-Walker spacetime for which we make a transformation d t = a ( t ) d τ d t = a ( t ) d τ dt=a(t)dtau\mathrm{d} t=a(t) \mathrm{d} \taudt=a(t)dτ leading to the conformal transformation d s 2 = a ( t ) 2 d s ¯ 2 d s 2 = a ( t ) 2 d s ¯ 2 ds^(2)=a(t)^(2)d bar(s)^(2)\mathrm{d} s^{2}=a(t)^{2} \mathrm{~d} \bar{s}^{2}ds2=a(t)2 ds¯2.
(a) Consider a solution to the electromagnetic wave equation of the form A μ e i B τ A μ e i B τ A_(mu)prope^(iB tau)A_{\mu} \propto \mathrm{e}^{\mathrm{i} B \tau}AμeiBτ, with B B BBB a constant, in the spacetime with line element d s ¯ d s ¯ d bar(s)\mathrm{d} \bar{s}ds¯. What form will this solution take in the spacetime with line element d s d s ds\mathrm{d} sds in terms of the variable t t ttt ?
(b) Explain why this implies that the instantaneous frequency of radiation ω ( t ) ω ( t ) omega(t)\omega(t)ω(t) obeys ω ( t ) a ( t ) = ω ( t ) a ( t ) = omega(t)a(t)=\omega(t) a(t)=ω(t)a(t)= const.
This equation is another demonstration of the cosmological redshift that occurs for light, emitted at time t 1 t 1 t_(1)t_{1}t1 and observed at time t 2 t 2 t_(2)t_{2}t2, for an expanding universe with a ( t 2 ) > a ( t 1 ) a t 2 > a t 1 a(t_(2)) > a(t_(1))a\left(t_{2}\right)>a\left(t_{1}\right)a(t2)>a(t1).
(19.6) What is the relationship between coordinates ( t , x ) ( t , x ) (t,x)(t, x)(t,x) and ( t , x t , x t^('),x^(')t^{\prime}, x^{\prime}t,x ) in the Penrose diagram for Minkowski space? Use this to justify the behaviour of the geodesics in Fig. 19.6.
(19.7) Consider the Penrose diagram for Minkowski space and those geodesics starting at the origin moving radially outwards.
(a) Draw the null geodesic followed by a light pulse.
(b) Draw the timelike geodesic followed by a particle with a speed β = 0.3 β = 0.3 beta=0.3\beta=0.3β=0.3.
(c) Draw the spacelike geodesic for β = 1.1 β = 1.1 beta=1.1\beta=1.1β=1.1.
(19.8) Consider the story told in Example 19.10. We shall analyse this using Sidney Coleman's method (explained further in Coleman, 2022).
(a) Show that the Friedmann equation predicts that during the evolution, we have
(19.39) d t = d a ( C / a 1 ) 1 2 (19.39) d t = d a ( C / a 1 ) 1 2 {:(19.39)dt=(da)/((C//a-1)^((1)/(2))):}\begin{equation*} \mathrm{d} t=\frac{\mathrm{d} a}{(C / a-1)^{\frac{1}{2}}} \tag{19.39} \end{equation*}(19.39)dt=da(C/a1)12
(b) Define the comoving distance η η eta\etaη via d t = a ( t ) d η d t = a ( t ) d η dt=a(t)deta\mathrm{d} t=a(t) \mathrm{d} \etadt=a(t)dη and show that a radial light ray travels a distance given by η η eta\etaη.
This comoving distance η η eta\etaη is identical to the conformal time coordinate in eqn 19.25.
(c) Show further that
(19.40) cos η = 1 2 a C (19.40) cos η = 1 2 a C {:(19.40)cos eta=1-(2a)/(C):}\begin{equation*} \cos \eta=1-\frac{2 a}{C} \tag{19.40} \end{equation*}(19.40)cosη=12aC
solves the equation of motion.
(d) Finally, show that, as parametrized by η η eta\etaη, we have
a ( η ) = C 2 ( 1 cos η ) (19.41) t ( η ) = C 2 ( η sin η ) a ( η ) = C 2 ( 1 cos η ) (19.41) t ( η ) = C 2 ( η sin η ) {:[a(eta)=(C)/(2)(1-cos eta)],[(19.41)t(eta)=(C)/(2)*(eta-sin eta)]:}\begin{align*} & a(\eta)=\frac{C}{2}(1-\cos \eta) \\ & t(\eta)=\frac{C}{2} \cdot(\eta-\sin \eta) \tag{19.41} \end{align*}a(η)=C2(1cosη)(19.41)t(η)=C2(ηsinη)
(e) Use this parametrization to verify the story in the example.

Part IV

Orbits, stars, and black holes

In Part IV of the book, we turn to the questions of some of the strongly gravitating objects in the Universe and the motions that they promote.
  • In Chapter 20, we review the machinery of Newtonian gravity and the methods for determining the possible motions in a Newtonian gravitating potential.
  • In Chapter 21, we discuss the Schwarzschild metric, which describes a stationary, spherically symmetric distribution of mass. We then determine the possible motions allowed by such a metric in Chapters 22, 23 and 24.
  • In Chapters 25-29, we examine static black holes, objects that gravitate so strongly that not even light can escape their interior. Black holes likely contain a singularity in spacetime, which are understood using tools from Chapters 26 and 27. In Chapter 28, we discuss black hole thermodynamics and the radiation that can emerge from black holes. Finally, in Chapter 29, we discuss the rich structures that result if we allow our black holes to carry charge or to rotate in space.

20

Newtonian orbits
20.1 Kepler's laws 219
20.2 Anatomy of an orbit 220
20.3 Effective potentials 222
20.4 Allowed trajectories 223
20.5 The why? of orbits 225
Chapter summary 227
Exercises 227
20.1 Kepler's laws 219 20.2 Anatomy of an orbit 220 20.3 Effective potentials 222 20.4 Allowed trajectories 223 20.5 The why? of orbits 225 Chapter summary 227 Exercises 227| 20.1 Kepler's laws | 219 | | :--- | :--- | | 20.2 Anatomy of an orbit | 220 | | 20.3 Effective potentials | 222 | | 20.4 Allowed trajectories | 223 | | 20.5 The why? of orbits | 225 | | Chapter summary | 227 | | Exercises | 227 |
1 1 ^(1){ }^{1}1 We shall temporarily restore factors of G G GGG in this chapter to make contact with familiar-looking equations.
2 2 ^(2){ }^{2}2 The dot denotes a derivative with re spect to t t ttt in this chapter.
3 3 ^(3){ }^{3}3 The particle is at r = r e r r = r e r vec(r)=r vec(e)_( vec(r))\vec{r}=r \vec{e}_{\vec{r}}r=rer and note that in polar coordinates
$$
e r ^ r = 0 , e r ^ θ = e θ ^ e θ ^ r = 0 , e θ ^ θ = e r ^ e r ^ r = 0 , e r ^ θ = e θ ^ e θ ^ r = 0 , e θ ^ θ = e r ^ {:[(del vec(e)_( hat(r)))/(del r)=0","quad(del vec(e)_( hat(r)))/(del theta)= vec(e)_( hat(theta))],[(del vec(e)_( hat(theta)))/(del r)=0","quad(del vec(e)_( hat(theta)))/(del theta)=- vec(e)_( hat(r))]:}\begin{aligned} & \frac{\partial \vec{e}_{\hat{r}}}{\partial r}=0, \quad \frac{\partial \vec{e}_{\hat{r}}}{\partial \theta}=\vec{e}_{\hat{\theta}} \\ & \frac{\partial \vec{e}_{\hat{\theta}}}{\partial r}=0, \quad \frac{\partial \vec{e}_{\hat{\theta}}}{\partial \theta}=-\vec{e}_{\hat{r}} \end{aligned}er^r=0,er^θ=eθ^eθ^r=0,eθ^θ=er^
w h e r e , a s a d v e r t i s e d i n E x a m p l e 10.4 , w e u s e o r t h o n o r m a l c o o r d i n a t e s i n t h i s p r o b l e m ( d e n o t e d b y h a t s o n i n d i c e s ) . H e n c e , w h e r e , a s a d v e r t i s e d i n E x a m p l e 10.4 , w e u s e o r t h o n o r m a l c o o r d i n a t e s i n t h i s p r o b l e m ( d e n o t e d b y h a t s o n i n d i c e s ) . H e n c e , where,asadvertisedinExample 10.4,weuseorthonormalcoordinatesinthisproblem(denotedbyhatsonindices).Hence,where, as advertised in Example 10.4, we use orthonormal coordinates in this problem (denoted by hats on indices). Hence,where,asadvertisedinExample10.4,weuseorthonormalcoordinatesinthisproblem(denotedbyhatsonindices).Hence,
\vec{v}=\dot{\vec{r}}=\dot{r} \vec{e}{\hat{r}}+r \dot{\vec{e}}{\hat{r}}=\dot{r} \vec{e}{\hat{r}}+r \dot{\theta} \vec{e}{\hat{\theta}} .
$$
The moon gravitates towards the earth and by the force of gravity is continually drawn off from a rectilinear motion and retained in its orbit.
Isaac Newton
'But the Solar System!' I protested.
'What the deuce is it to me?' (Sherlock Holmes) interrupted impatiently: 'you say that we go round the sun. If we went round the moon it would not make a penny-worth of difference to me and my work.'
Arthur Conan Doyle (1859-1930) A Study in Scarlet
General relativity is well known to provide corrections to Newtonian gravitation. In this chapter, we take a step back, and present a set of methods to find and describe the trajectories allowed by the Φ ( r ) 1 / r Φ ( r ) 1 / r Phi(r)prop-1//r\Phi(r) \propto-1 / rΦ(r)1/r potential of Newtonian gravitation. On solving the problem we shall find that there is a restricted class of possible trajectories. It turns out that the same methods employed here can be used to deal with the more varied trajectories allowed in general relativity.
In Newtonian gravitation, the potential energy due to the gravitational interaction between two particles with masses m m mmm and M M MMM, separated by a distance r r rrr, is given by 1 U ( r ) = G M m / r 1 U ( r ) = G M m / r ^(1)U(r)=-GMm//r{ }^{1} U(r)=-G M m / r1U(r)=GMm/r. We take the mass M M MMM to be fixed at the origin and the mass m m mmm to be separated from M M MMM by the 3 -vector r r vec(r)\vec{r}r, and moving with momentum p = m v p = m v vec(p)=m vec(v)\vec{p}=m \vec{v}p=mv. The Newtonian force on the mass m m mmm, given by F ( r ) = G M m r / r 3 F ( r ) = G M m r / r 3 vec(F)(r)=-GMm vec(r)//r^(3)\vec{F}(r)=-G M m \vec{r} / r^{3}F(r)=GMmr/r3, acts radially, and therefore does not give rise to any torque τ τ vec(tau)\vec{\tau}τ. This is because τ = r × F τ = r × F vec(tau)= vec(r)xx vec(F)\vec{\tau}=\vec{r} \times \vec{F}τ=r×F and so, for a central force such as Newtonian gravitation, τ r × r = 0 τ r × r = 0 vec(tau)prop vec(r)xx vec(r)=0\vec{\tau} \propto \vec{r} \times \vec{r}=0τr×r=0. Since the angular momentum of the moving mass L = r × p L = r × p vec(L)= vec(r)xx vec(p)\vec{L}=\vec{r} \times \vec{p}L=r×p is related to the torque via 2 τ = L ˙ 2 τ = L ˙ ^(2) vec(tau)= vec(L)^(˙){ }^{2} \vec{\tau}=\dot{\vec{L}}2τ=L˙, the angular momentum for any central force is a constant of the motion. This constant angular momentum can be used, along with the constant energy of the system E E EEE, to classify the possible trajectories of the moving particle.
As a further result of the conservation of angular momentum, the vector L L vec(L)\vec{L}L (which is, by definition, perpendicular to r r vec(r)\vec{r}r and p p vec(p)\vec{p}p ) remains fixed. This means that the path of a particle in Newtonian gravitation can be taken as being confined to the equatorial plane of a set of threedimensional spherical coordinates. Taking L L vec(L)\vec{L}L to be pointing along the z z zzz-direction, we can therefore follow the paths using the two-dimensional cylindrical coordinates ( r , θ ) ( r , θ ) (r,theta)(r, \theta)(r,θ). In these polar coordinates, we have 3 3 ^(3){ }^{3}3 velocity components v r ^ = r ˙ v r ^ = r ˙ v_( hat(r))=r^(˙)v_{\hat{r}}=\dot{r}vr^=r˙ and v θ ^ = r θ ˙ v θ ^ = r θ ˙ v_( hat(theta))=rtheta^(˙)v_{\hat{\theta}}=r \dot{\theta}vθ^=rθ˙ and hence a squared velocity
v 2 = ( r ˙ 2 + r 2 θ ˙ 2 ) v 2 = r ˙ 2 + r 2 θ ˙ 2 v^(2)=(r^(˙)^(2)+r^(2)theta^(˙)^(2))v^{2}=\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)v2=(r˙2+r2θ˙2). The system has a Lagrangian
(20.1) L = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) + G M m r . (20.1) L = 1 2 m r ˙ 2 + r 2 θ ˙ 2 + G M m r . {:(20.1)L=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))+(GMm)/(r).:}\begin{equation*} L=\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)+\frac{G M m}{r} . \tag{20.1} \end{equation*}(20.1)L=12m(r˙2+r2θ˙2)+GMmr.
Feeding this into the Euler-Lagrange equation, we obtain the two equations of motion 4 4 ^(4){ }^{4}4
r ¨ r θ ˙ 2 = G M r 2 , (20.5) r θ ¨ + 2 r ˙ θ ˙ = 0 . r ¨ r θ ˙ 2 = G M r 2 , (20.5) r θ ¨ + 2 r ˙ θ ˙ = 0 . {:[r^(¨)-rtheta^(˙)^(2)=-(GM)/(r^(2))","],[(20.5)rtheta^(¨)+2r^(˙)theta^(˙)=0.]:}\begin{align*} \ddot{r}-r \dot{\theta}^{2} & =-\frac{G M}{r^{2}}, \\ r \ddot{\theta}+2 \dot{r} \dot{\theta} & =0 . \tag{20.5} \end{align*}r¨rθ˙2=GMr2,(20.5)rθ¨+2r˙θ˙=0.
We will meet some solutions to these equations in this chapter.

20.1 Kepler's laws

An orbit is the closed, periodic trajectory that each planet in our Solar System is, to a good approximation, observed to follow. Before Isaac Newton provided a solution to the problem of the description of orbits, Johannes Kepler 5 5 ^(5){ }^{5}5 had formulated three laws of planetary motion resulting from his analysis of the observed trajectories of planets in our solar system. These laws are useful in understanding what the allowed orbits are. Each can be proven using the Newtonian approach, and we shall do that here. The laws are given below.
Kepler's first law: Planetary orbits follow an elliptical trajectory, with the Sun at a focus of the ellipse.
Kepler's second law: The line from Sun to planet sweeps out equal areas in equal times.
Kepler's third law: The square of the period of an orbit is proportional to the cube of the semi-major axis of the ellipse that describes its orbit.
We can immediately show how Kepler's second law is a consequence of the conservation of angular momentum.
Example 20.1
Since there is no component of the Newtonian force in the θ θ theta\thetaθ direction, we have, from eqn 20.4, that
(20.6) r θ ¨ + 2 r ˙ θ ˙ = 0 . (20.7) r 2 θ ˙ = C , (20.6) r θ ¨ + 2 r ˙ θ ˙ = 0 . (20.7) r 2 θ ˙ = C , {:[(20.6)rtheta^(¨)+2r^(˙)theta^(˙)=0.],[(20.7)r^(2)theta^(˙)=C","]:}\begin{gather*} r \ddot{\theta}+2 \dot{r} \dot{\theta}=0 . \tag{20.6}\\ r^{2} \dot{\theta}=C, \tag{20.7} \end{gather*}(20.6)rθ¨+2r˙θ˙=0.(20.7)r2θ˙=C,
Integrating, we find
where C C CCC is a constant. As shown in Fig. 20.1, C C CCC can be interpreted geometrically as twice the rate at which the radius vector sweeps out area A A AAA. We therefore have
(20.8) d A d t = 1 2 r 2 d θ d t = const. (20.8) d A d t = 1 2 r 2 d θ d t =  const.  {:(20.8)(dA)/((d)t)=(1)/(2)r^(2)((d)theta)/((d)t)=" const. ":}\begin{equation*} \frac{\mathrm{d} A}{\mathrm{~d} t}=\frac{1}{2} r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}=\text { const. } \tag{20.8} \end{equation*}(20.8)dA dt=12r2 dθ dt= const. 
This is Kepler's second law (i.e. constant A ˙ A ˙ A^(˙)\dot{A}A˙ ), proven from the equations of motion. As a bonus, we notice that the angular momentum is given by
(20.9) L = r m v θ ^ e z ~ = m r 2 θ ˙ e Σ ~ (20.9) L = r m v θ ^ e z ~ = m r 2 θ ˙ e Σ ~ {:(20.9) vec(L)=rmv_( hat(theta)) vec(e)_( tilde(z))=mr^(2)theta^(˙) vec(e)_( tilde(Sigma)):}\begin{equation*} \vec{L}=r m v_{\hat{\theta}} \vec{e}_{\tilde{z}}=m r^{2} \dot{\theta} \vec{e}_{\tilde{\Sigma}} \tag{20.9} \end{equation*}(20.9)L=rmvθ^ez~=mr2θ˙eΣ~
4 4 ^(4){ }^{4}4 The components of acceleration can be derived using
a = v ˙ = r ¨ e r ^ + r ˙ e ˙ r ^ + r ˙ θ ˙ e θ ^ + r ˙ θ ¨ e θ ^ + r θ ˙ e ˙ θ ^ a = v ˙ = r ¨ e r ^ + r ˙ e ˙ r ^ + r ˙ θ ˙ e θ ^ + r ˙ θ ¨ e θ ^ + r θ ˙ e ˙ θ ^ vec(a)= vec(v)^(˙)=r^(¨) vec(e)_( hat(r))+r^(˙) vec(e)^(˙)_( hat(r))+r^(˙)theta^(˙) vec(e)_( hat(theta))+r^(˙)theta^(¨) vec(e)_( hat(theta))+rtheta^(˙) vec(e)^(˙)_( hat(theta))\vec{a}=\dot{\vec{v}}=\ddot{r} \vec{e}_{\hat{r}}+\dot{r} \dot{\vec{e}}_{\hat{r}}+\dot{r} \dot{\theta} \vec{e}_{\hat{\theta}}+\dot{r} \ddot{\theta} \vec{e}_{\hat{\theta}}+r \dot{\theta} \dot{\vec{e}}_{\hat{\theta}}a=v˙=r¨er^+r˙e˙r^+r˙θ˙eθ^+r˙θ¨eθ^+rθ˙e˙θ^
(20.2) = a r ^ e r ^ + a θ ^ e θ ^ (20.2) = a r ^ e r ^ + a θ ^ e θ ^ {:(20.2)=a_( hat(r)) vec(e)_( hat(r))+a_( hat(theta)) vec(e)_( hat(theta)):}\begin{equation*} =a_{\hat{r}} \vec{e}_{\hat{r}}+a_{\hat{\theta}} \vec{e}_{\hat{\theta}} \tag{20.2} \end{equation*}(20.2)=ar^er^+aθ^eθ^
where
(20.3) a r ^ = r ¨ r θ ˙ 2 (20.4) a θ ^ = r θ ¨ + 2 r ˙ θ ˙ . (20.3) a r ^ = r ¨ r θ ˙ 2 (20.4) a θ ^ = r θ ¨ + 2 r ˙ θ ˙ . {:[(20.3)a_( hat(r))=r^(¨)-rtheta^(˙)^(2)],[(20.4)a_( hat(theta))=rtheta^(¨)+2r^(˙)theta^(˙).]:}\begin{align*} & a_{\hat{r}}=\ddot{r}-r \dot{\theta}^{2} \tag{20.3}\\ & a_{\hat{\theta}}=r \ddot{\theta}+2 \dot{r} \dot{\theta} . \tag{20.4} \end{align*}(20.3)ar^=r¨rθ˙2(20.4)aθ^=rθ¨+2r˙θ˙.
This means that eqn 20.5 is a statement of F = m a F = m a vec(F)=m vec(a)\vec{F}=m \vec{a}F=ma with m m mmm divided out.
5 5 ^(5){ }^{5}5 Johannes Kepler (1571-1630) was plagued by ill health, complaining of myopia, multiple vision, sores, stomach and gall bladder problems, piles, rashes, mange, worms, and the delusion that he was a dog. At one point he had to defend his mother against charges of witchcraft. His work on orbits was motivated by his attempts to describe the planetary orbits in terms of plantonic solids, using data from Tycho Brahe's (1546-1601) naked-eye observations. The problem of computing the orbits of a single particle under Newtonian gravity continues to be called the Kepler problem.
Fig. 20.1 A small displacement to the radius vector r ( t ) r ( t ) vec(r)(t)\vec{r}(t)r(t) changes the vector by an amount | Δ r | = 2 r sin Δ θ / 2 r Δ θ | Δ r | = 2 r sin Δ θ / 2 r Δ θ |Delta vec(r)|=2r sin Delta theta//2~~r Delta theta|\Delta \vec{r}|=2 r \sin \Delta \theta / 2 \approx r \Delta \theta|Δr|=2rsinΔθ/2rΔθ. The area of the triangle in the figure is then Δ A = 1 2 r 2 Δ θ Δ A = 1 2 r 2 Δ θ Delta A=(1)/(2)r^(2)Delta theta\Delta A=\frac{1}{2} r^{2} \Delta \thetaΔA=12r2Δθ, and so area is swept out at a rate A ˙ = r 2 θ ˙ / 2 ( = C / 2 ) A ˙ = r 2 θ ˙ / 2 ( = C / 2 ) A^(˙)=r^(2)theta^(˙)//2(=C//2)\dot{A}=r^{2} \dot{\theta} / 2(=C / 2)A˙=r2θ˙/2(=C/2).
Fig. 20.2 The ellipse is a conic section, achieved via a slice made at an angle to the base. A slice made parallel to the
Fig. 20.3 The anatomy of an ellipse, showing the foci F F FFF and F F F^(')F^{\prime}F, semi-major axis a a aaa, the semi-minor axis b b bbb and the eccentricity ϵ ϵ epsilon\epsilonϵ. The point P P P\mathcal{P}P is a distance r r rrr from F F FFF. A circular orbit is a special case where a = b , ϵ = 0 a = b , ϵ = 0 a=b,epsilon=0a=b, \epsilon=0a=b,ϵ=0 and F F FFF coincides with F F F^(')F^{\prime}F.
Taking the area to be a vector A A vec(A)\vec{A}A parallel to e z ~ e z ~ vec(e)_( tilde(z))\vec{e}_{\tilde{z}}ez~, we have
(20.10) d A d t = 1 2 r 2 d θ d t e z ^ = L 2 m (20.10) d A d t = 1 2 r 2 d θ d t e z ^ = L 2 m {:(20.10)(d( vec(A)))/((d)t)=(1)/(2)r^(2)((d)theta)/((d)t) vec(e)_( hat(z))=(( vec(L)))/(2m):}\begin{equation*} \frac{\mathrm{d} \vec{A}}{\mathrm{~d} t}=\frac{1}{2} r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} t} \vec{e}_{\hat{z}}=\frac{\vec{L}}{2 m} \tag{20.10} \end{equation*}(20.10)dA dt=12r2 dθ dtez^=L2m
Since L L vec(L)\vec{L}L is a constant, A ˙ A ˙ vec(A)^(˙)\dot{\vec{A}}A˙ is also constant. Kepler's second law therefore reflects the conservation of angular momentum, which itself follows for any central gravitational force.

20.2 Anatomy of an orbit

Kepler's first law states that orbits follow elliptical paths. The description of orbits enjoys a rich range of terminology and here we review the necessary vocabulary.
An ellipse is one of a class of two-dimensional figures known as the conic sections that also includes the circle, the parabola and the hyperbola. As the name suggests, these are figures that are generated by taking slices of cones. The slice through a cone needed to generate an ellipse is shown in Fig. 20.2 and a typical ellipse is shown in Fig. 20.3. An ellipse is described by an equation in polar coordinates
(20.11) 1 r = a b 2 ( 1 + ϵ cos θ ) , 0 ϵ < 1 , (20.11) 1 r = a b 2 ( 1 + ϵ cos θ ) , 0 ϵ < 1 , {:(20.11)(1)/(r)=(a)/(b^(2))(1+epsilon cos theta)","quad0 <= epsilon < 1",":}\begin{equation*} \frac{1}{r}=\frac{a}{b^{2}}(1+\epsilon \cos \theta), \quad 0 \leq \epsilon<1, \tag{20.11} \end{equation*}(20.11)1r=ab2(1+ϵcosθ),0ϵ<1,
where a a aaa and b b bbb are, respectively, the semi-major and semi-minor axes of the ellipse and ϵ ϵ epsilon\epsilonϵ is known as the eccentricity. [The angle θ θ theta\thetaθ is the one that the radial vector r r vec(r)\vec{r}r (linking the planet's position P P P\mathcal{P}P and the focus) makes to the radial vector corresponding to the planet being at its position of closest approach to the focus, as shown in Fig. 20.3.] With an equation for the elliptical trajectory available, we can show that it is compatible with a 1 / r 2 1 / r 2 1//r^(2)1 / r^{2}1/r2 central force.

Example 20.2

We shall investigate the nature of the acceleration of a particle following an elliptical trajectory. Differentiate the equation for an ellipse to find
(20.12) r ˙ r 2 = ϵ a b 2 sin θ θ ˙ (20.12) r ˙ r 2 = ϵ a b 2 sin θ θ ˙ {:(20.12)((r^(˙)))/(r^(2))=(epsilon a)/(b^(2))sin thetatheta^(˙):}\begin{equation*} \frac{\dot{r}}{r^{2}}=\frac{\epsilon a}{b^{2}} \sin \theta \dot{\theta} \tag{20.12} \end{equation*}(20.12)r˙r2=ϵab2sinθθ˙
We have r 2 θ ˙ = C r 2 θ ˙ = C r^(2)theta^(˙)=Cr^{2} \dot{\theta}=Cr2θ˙=C, where C C CCC is a constant equal to | L | / m | L | / m | vec(L)|//m|\vec{L}| / m|L|/m. Substituting this gives
(20.13) r ˙ = C ϵ a b 2 sin θ (20.13) r ˙ = C ϵ a b 2 sin θ {:(20.13)r^(˙)=(C epsilon a)/(b^(2))sin theta:}\begin{equation*} \dot{r}=\frac{C \epsilon a}{b^{2}} \sin \theta \tag{20.13} \end{equation*}(20.13)r˙=Cϵab2sinθ
and then
(20.14) r ¨ = C ϵ a b 2 cos θ θ ˙ = r ¨ r θ ˙ 2 we find (20.14) r ¨ = C ϵ a b 2 cos θ θ ˙ = r ¨ r θ ˙ 2  we find  {:[(20.14)r^(¨)=(C epsilon a)/(b^(2))*cos thetatheta^(˙)],[=r^(¨)-rtheta^(˙)^(2)" we find "]:}\begin{gather*} \ddot{r}=\frac{C \epsilon a}{b^{2}} \cdot \cos \theta \dot{\theta} \tag{20.14}\\ =\ddot{r}-r \dot{\theta}^{2} \text { we find } \end{gather*}(20.14)r¨=Cϵab2cosθθ˙=r¨rθ˙2 we find 
(20.15) a r ^ = C 2 r 2 ( ϵ a cos θ b 2 1 r ) (20.15) a r ^ = C 2 r 2 ϵ a cos θ b 2 1 r {:(20.15)a_( hat(r))=(C^(2))/(r^(2))((epsilon a cos theta)/(b^(2))-(1)/(r)):}\begin{equation*} a_{\hat{r}}=\frac{C^{2}}{r^{2}}\left(\frac{\epsilon a \cos \theta}{b^{2}}-\frac{1}{r}\right) \tag{20.15} \end{equation*}(20.15)ar^=C2r2(ϵacosθb21r)
Using the radial acceleration a r ^ = r ¨ r θ ˙ 2 a r ^ = r ¨ r θ ˙ 2 a_( hat(r))=r^(¨)-rtheta^(˙)^(2)a_{\hat{r}}=\ddot{r}-r \dot{\theta}^{2}ar^=r¨rθ˙2 we find
Referring back to the equation for an ellipse we see that the part in the bracket is equal to 6 a / b 2 6 a / b 2 ^(6)-a//b^(2){ }^{6}-a / b^{2}6a/b2 and so we can write
(20.16) a r ^ = C 2 a b 2 1 r 2 (20.16) a r ^ = C 2 a b 2 1 r 2 {:(20.16)a_( hat(r))=-(C^(2)a)/(b^(2))(1)/(r^(2)):}\begin{equation*} a_{\hat{r}}=-\frac{C^{2} a}{b^{2}} \frac{1}{r^{2}} \tag{20.16} \end{equation*}(20.16)ar^=C2ab21r2
Since acceleration is proportional to force, we conclude that an elliptical orbit is compatible with a 1 / r 2 1 / r 2 1//r^(2)1 / r^{2}1/r2 force law.
The expression for radial acceleration allows us to prove Kepler's third law.

Example 20.3

The area of an ellipse is given by A = π a b A = π a b A=pi abA=\pi a bA=πab. Using the notation from the previous example, the period of the orbit is T = π a b / ( C / 2 ) T = π a b / ( C / 2 ) T=pi ab//(C//2)T=\pi a b /(C / 2)T=πab/(C/2). Substituting into eqn 20.16 we have
(20.17) a r ^ = 4 π 2 a 3 T 2 1 r 2 . (20.17) a r ^ = 4 π 2 a 3 T 2 1 r 2 . {:(20.17)a_( hat(r))=-(4pi^(2)a^(3))/(T^(2))(1)/(r^(2)).:}\begin{equation*} a_{\hat{r}}=-\frac{4 \pi^{2} a^{3}}{T^{2}} \frac{1}{r^{2}} . \tag{20.17} \end{equation*}(20.17)ar^=4π2a3T21r2.
We also know that a r ^ = G M / r 2 a r ^ = G M / r 2 a_( hat(r))=-GM//r^(2)a_{\hat{r}}=-G M / r^{2}ar^=GM/r2 and upon equating these expressions we have
(20.18) T 2 = 4 π 2 a 3 G M (20.18) T 2 = 4 π 2 a 3 G M {:(20.18)T^(2)=(4pi^(2)a^(3))/(GM):}\begin{equation*} T^{2}=\frac{4 \pi^{2} a^{3}}{G M} \tag{20.18} \end{equation*}(20.18)T2=4π2a3GM
which shows that the square of the period is proportional to the cube of a a aaa, the semi-major axis of the ellipse.
For an object in orbit around the Sun, the perihelion is the position of closest approach to the Sun. The aphelion is the position in the orbit furthest from the Sun (Fig. 20.4). These words are often incorrectly applied to the orbits of objects around bodies other than the Sun. In fact, for (i) orbits around the Earth the points are called the perigee and apogee; (ii) for orbits around a star we call them the periastron and apastron and (iii), most generally, for orbits around any centre of mass, they are called the periapsis and apoapsis.
We can exploit the properties of the aphelion to relate the total energy of the orbiting planet to the dimensions of the ellipse.

Example 20.4

At the aphelion we have r = a ( 1 + ϵ ) r = a ( 1 + ϵ ) r=a(1+epsilon)r=a(1+\epsilon)r=a(1+ϵ). Also, since at this point θ = π θ = π theta=pi\theta=\piθ=π and so, from eqn 20.11, b 2 = a 2 ( 1 ϵ 2 ) b 2 = a 2 1 ϵ 2 b^(2)=a^(2)(1-epsilon^(2))b^{2}=a^{2}\left(1-\epsilon^{2}\right)b2=a2(1ϵ2). The potential energy at the aphelion for a planet in orbit around the Sun is
(20.19) U = G M m a ( 1 + ϵ ) (20.19) U = G M m a ( 1 + ϵ ) {:(20.19)U=-(GMm)/(a(1+epsilon)):}\begin{equation*} U=-\frac{G M m}{a(1+\epsilon)} \tag{20.19} \end{equation*}(20.19)U=GMma(1+ϵ)
At this point v r ^ = 0 v r ^ = 0 v_( hat(r))=0v_{\hat{r}}=0vr^=0 and so the kinetic energy is
(20.20) K = 1 2 m v θ ^ 2 = 1 2 m r 2 θ ˙ 2 = m C 2 2 r 2 (20.20) K = 1 2 m v θ ^ 2 = 1 2 m r 2 θ ˙ 2 = m C 2 2 r 2 {:(20.20)K=(1)/(2)mv_( hat(theta))^(2)=(1)/(2)mr^(2)theta^(˙)^(2)=(mC^(2))/(2r^(2)):}\begin{equation*} K=\frac{1}{2} m v_{\hat{\theta}}^{2}=\frac{1}{2} m r^{2} \dot{\theta}^{2}=\frac{m C^{2}}{2 r^{2}} \tag{20.20} \end{equation*}(20.20)K=12mvθ^2=12mr2θ˙2=mC22r2
where we've used eqn 20.7 in the final step. Substituting for r r rrr we find
(20.21) K = m C 2 2 a 2 ( 1 + ϵ ) 2 (20.21) K = m C 2 2 a 2 ( 1 + ϵ ) 2 {:(20.21)K=(mC^(2))/(2a^(2)(1+epsilon)^(2)):}\begin{equation*} K=\frac{m C^{2}}{2 a^{2}(1+\epsilon)^{2}} \tag{20.21} \end{equation*}(20.21)K=mC22a2(1+ϵ)2
or, substituting 7 7 ^(7){ }^{7}7 for C C CCC, we obtain
(20.23) K = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) (20.23) K = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) {:(20.23)K=(GMm(1-epsilon))/(2a(1+epsilon)):}\begin{equation*} K=\frac{G M m(1-\epsilon)}{2 a(1+\epsilon)} \tag{20.23} \end{equation*}(20.23)K=GMm(1ϵ)2a(1+ϵ)
This allows us to sum
(20.24) E = K + U = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) G M m a ( 1 + ϵ ) , (20.25) E = G M m 2 a , (20.24) E = K + U = G M m ( 1 ϵ ) 2 a ( 1 + ϵ ) G M m a ( 1 + ϵ ) , (20.25) E = G M m 2 a , {:[(20.24)E=K+U=(GMm(1-epsilon))/(2a(1+epsilon))-(GMm)/(a(1+epsilon))","],[(20.25)E=-(GMm)/(2a)","]:}\begin{gather*} E=K+U=\frac{G M m(1-\epsilon)}{2 a(1+\epsilon)}-\frac{G M m}{a(1+\epsilon)}, \tag{20.24}\\ E=-\frac{G M m}{2 a}, \tag{20.25} \end{gather*}(20.24)E=K+U=GMm(1ϵ)2a(1+ϵ)GMma(1+ϵ),(20.25)E=GMm2a,
Fig. 20.4 The perihelion p p ppp and aphelion a a aaa of an orbit around the Sun S at one of the foci of the elliptical orbit.
7 7 ^(7){ }^{7}7 Write
C 2 = 4 π 2 a 2 b 2 T 2 = 4 π 2 a 4 ( 1 ϵ 2 ) T 2 C 2 = 4 π 2 a 2 b 2 T 2 = 4 π 2 a 4 1 ϵ 2 T 2 C^(2)=(4pi^(2)a^(2)b^(2))/(T^(2))=(4pi^(2)a^(4)(1-epsilon^(2)))/(T^(2))C^{2}=\frac{4 \pi^{2} a^{2} b^{2}}{T^{2}}=\frac{4 \pi^{2} a^{4}\left(1-\epsilon^{2}\right)}{T^{2}}C2=4π2a2b2T2=4π2a4(1ϵ2)T2
and use Kepler's third law
T 2 = 4 π 2 a 3 G M T 2 = 4 π 2 a 3 G M T^(2)=(4pi^(2)a^(3))/(GM)T^{2}=\frac{4 \pi^{2} a^{3}}{G M}T2=4π2a3GM
Since | L | = C m | L | = C m |L|=Cm|L|=C m|L|=Cm, we can use this to deive the useful result that
L 2 = G M m 2 a ( 1 ϵ 2 ) , L 2 = G M m 2 a 1 ϵ 2 , L^(2)=GMm^(2)a(1-epsilon^(2)),L^{2}=G M m^{2} a\left(1-\epsilon^{2}\right),L2=GMm2a(1ϵ2),
for an elliptical orbit.
Fig. 20.5 The two contributions to the Newtonian effective potential energy (the repulsive angular momentum barrier and attractive 1 / r 1 / r -1//r-1 / r1/r contribution, shown by dotted lines) give the curve shown with the solid line.
8 8 ^(8){ }^{8}8 This term is sometimes called the angular momentum barrier.
9 9 ^(9){ }^{9}9 It lies at the minimum because the system's energy is determined only by U eff U eff  U_("eff ")U_{\text {eff }}Ueff .
Fig. 20.6 Energies for an elliptical orbit with energy E 1 < 0 E 1 < 0 E_(1) < 0E_{1}<0E1<0 and an unbound trajectory ( E 2 > 0 ) E 2 > 0 (E_(2) > 0)\left(E_{2}>0\right)(E2>0).

20.3 Effective potentials

The method of effective potentials is especially useful in understanding which trajectories are possible. We shall use it in the relativistic case and so we introduce it here. If the particle of mass m m mmm is moving with velocity v v vec(v)\vec{v}v in the field of the stationary mass M M MMM then the energy of the two-particle system is written as
(20.26) E = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) G M m r (20.26) E = 1 2 m r ˙ 2 + r 2 θ ˙ 2 G M m r {:(20.26)E=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))-(GMm)/(r):}\begin{equation*} E=\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)-\frac{G M m}{r} \tag{20.26} \end{equation*}(20.26)E=12m(r˙2+r2θ˙2)GMmr
The angular momentum has a magnitude | L | = m θ ˙ r 2 | L | = m θ ˙ r 2 |L|=mtheta^(˙)r^(2)|L|=m \dot{\theta} r^{2}|L|=mθ˙r2, which allows us to substitute for θ ˙ θ ˙ theta^(˙)\dot{\theta}θ˙ and then drop the angular contribution into an effective potential energy function U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r). Specifically, we write
(20.27) E = 1 2 m r ˙ 2 + U eff ( r ) (20.27) E = 1 2 m r ˙ 2 + U eff ( r ) {:(20.27)E=(1)/(2)mr^(˙)^(2)+U_(eff)(r):}\begin{equation*} E=\frac{1}{2} m \dot{r}^{2}+U_{\mathrm{eff}}(r) \tag{20.27} \end{equation*}(20.27)E=12mr˙2+Ueff(r)
where the effective potential energy is given by
(20.28) U eff ( r ) = L 2 2 m r 2 G M m r (20.28) U eff ( r ) = L 2 2 m r 2 G M m r {:(20.28)U_(eff)(r)=(L^(2))/(2mr^(2))-(GMm)/(r):}\begin{equation*} U_{\mathrm{eff}}(r)=\frac{L^{2}}{2 m r^{2}}-\frac{G M m}{r} \tag{20.28} \end{equation*}(20.28)Ueff(r)=L22mr2GMmr
Given an initial angular momentum of a particle in this field, we can write the effective potential. As we shall see, it is this effective potential that can be used to classify and understand all of the trajectories that are possible for particles in the Newtonian potential. The potential energy function U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r) has the two contributions shown in Fig. 20.5. There is (i) a repulsive contribution whose strength depends on the angular momentum, 8 8 ^(8){ }^{8}8 going as 1 / r 2 1 / r 2 1//r^(2)1 / r^{2}1/r2; and (ii) an attractive contribution, given by the usual 1 / r 1 / r -1//r-1 / r1/r potential. The difference in power law and sign means that the resultant potential can have a minimum. The limits of the effective potential for small and large r r rrr are instructive too: (i) the effect of gravitation dies off at large distances as 1 / r 1 / r 1//r1 / r1/r irrespective of the angular momentum; (ii) the potential gets very large at small r r rrr owing to the angular momentum term. This means that any particle with non-zero angular momentum is scattered by the potential. (There is no option to spiral into the origin, for example). The only trajectory that can end up at the origin has | L | = 0 | L | = 0 |L|=0|L|=0|L|=0, which amounts to a radial plunge into the source of the potential.
The values of r r rrr for which E = U eff ( r ) E = U eff ( r ) E=U_(eff)(r)E=U_{\mathrm{eff}}(r)E=Ueff(r) set some limits on the motion. We have this condition when r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0. This doesn't mean that the particle has stopped (because we can still have θ ˙ 0 θ ˙ 0 theta^(˙)!=0\dot{\theta} \neq 0θ˙0 ), rather, the particle is turning at constant r r rrr, as it does at the points nearest and furthest from the focus of the ellipse. If r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0 at all times, then the trajectory must be circular (because there is never a change in r r rrr ) with a radius r 0 r 0 r_(0)r_{0}r0 that corresponds to the minimum 9 9 ^(9){ }^{9}9 of U eff U eff  U_("eff ")U_{\text {eff }}Ueff . This occurs when U eff ( r ) / r | r = r 0 = 0 U eff  ( r ) / r r = r 0 = 0 delU_("eff ")(r)// del r|_(r=r_(0))=0\partial U_{\text {eff }}(r) /\left.\partial r\right|_{r=r_{0}}=0Ueff (r)/r|r=r0=0 or
(20.29) L 2 m r 0 = G M m (circular orbit ) (20.29) L 2 m r 0 = G M m  (circular orbit  {:(20.29){:(L^(2))/(mr_(0))=GMm quad" (circular orbit "):}\begin{equation*} \left.\frac{L^{2}}{m r_{0}}=G M m \quad \text { (circular orbit }\right) \tag{20.29} \end{equation*}(20.29)L2mr0=GMm (circular orbit )
The circular orbit has a negative total energy E = G M m / 2 r 0 E = G M m / 2 r 0 E=-GMm//2r_(0)E=-G M m / 2 r_{0}E=GMm/2r0, so that the orbiting particle is in a bound state that requires energy to
be inputted in order to escape. There are other possible bound orbits with E < 0 E < 0 E < 0E<0E<0, where the particle can be thought of as being confined by the effective potential, bouncing back and forth between the two values where E = U eff 10 E = U eff  10 E=U_("eff ")^(10)E=U_{\text {eff }}{ }^{10}E=Ueff 10 These are, of course, the ellipses that Kepler's first law describes. We can immediately gain some insight by looking at the negative values of U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r), since E = U eff E = U eff  E=U_("eff ")E=U_{\text {eff }}E=Ueff  for the elliptical paths at the perihelion and aphelion, where r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0. As shown in Fig. 20.6, lines of negative constant total energy E E EEE intersect U eff ( r ) U eff  ( r ) U_("eff ")(r)U_{\text {eff }}(r)Ueff (r) at two points when r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0, providing the length of semi-major and semi-minor axes.
In contrast, an unbounded trajectory (i.e. not an orbit) with a positive energy has a unique distance of closest approach to the gravitating mass. At this position r ˙ = 0 r ˙ = 0 r^(˙)=0\dot{r}=0r˙=0 and so the distance of closest approach occurs when the energy of the system E = U eff ( r ) E = U eff  ( r ) E=U_("eff ")(r)E=U_{\text {eff }}(r)E=Ueff (r). We can therefore use the graph of the effective potential energy (Fig. 20.6) to describe all of the trajectories. We imagine the particle moving horizontally as its value of r r rrr varies; it is deflected whenever it meets the curve U eff U eff  U_("eff ")U_{\text {eff }}Ueff . Remember that the form of U eff U eff  U_("eff ")U_{\text {eff }}Ueff  is determined by the (constant) angular momentum of the particle. We shall use this graphical method again when we look into the trajectories allowed by general relativity.

20.4 Allowed trajectories

We can now solve the problem once and for all, by determining all of the possible motions in the Newtonian potential. We have seen that orbits are expected in a Newtonian potential and that these have a negative total energy E E EEE. If the energy is positive, then we expect the trajectory to be unbounded. We shall now solve the equations of motion for the trajectories of the particles and then use the sign of the total energy to classify the orbits. The technique of choice employs the variable u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r, which we shall also use in solving the relativistic problem.

Example 20.5

What is the equation of motion for the trajectories? It's convenient to use u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r and write
(20.30) d r d t = 1 u 2 d u d t = L m d u d θ (20.31) d 2 r d t 2 = L m d 2 u d θ 2 d θ d t = L 2 m 2 u 2 d 2 u d θ 2 (20.30) d r d t = 1 u 2 d u d t = L m d u d θ (20.31) d 2 r d t 2 = L m d 2 u d θ 2 d θ d t = L 2 m 2 u 2 d 2 u d θ 2 {:[(20.30)(dr)/((d)t)=-(1)/(u^(2))((d)u)/((d)t)=-(L)/(m)((d)u)/((d)theta)],[(20.31)(d^(2)r)/((d)t^(2))=-(L)/(m)(d^(2)u)/((d)theta^(2))((d)theta)/((d)t)=-(L^(2))/(m^(2))u^(2)(d^(2)u)/((d)theta^(2))]:}\begin{gather*} \frac{\mathrm{d} r}{\mathrm{~d} t}=-\frac{1}{u^{2}} \frac{\mathrm{~d} u}{\mathrm{~d} t}=-\frac{L}{m} \frac{\mathrm{~d} u}{\mathrm{~d} \theta} \tag{20.30}\\ \frac{\mathrm{~d}^{2} r}{\mathrm{~d} t^{2}}=-\frac{L}{m} \frac{\mathrm{~d}^{2} u}{\mathrm{~d} \theta^{2}} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}=-\frac{L^{2}}{m^{2}} u^{2} \frac{\mathrm{~d}^{2} u}{\mathrm{~d} \theta^{2}} \tag{20.31} \end{gather*}(20.30)dr dt=1u2 du dt=Lm du dθ(20.31) d2r dt2=Lm d2u dθ2 dθ dt=L2m2u2 d2u dθ2
and
Newton's second law becomes
F r ^ = G M m u 2 = m [ d 2 r d t 2 r ( d θ d t ) 2 ] (20.32) = L 2 m u 2 ( d 2 u d θ 2 + u ) F r ^ = G M m u 2 = m d 2 r d t 2 r d θ d t 2 (20.32) = L 2 m u 2 d 2 u d θ 2 + u {:[F_( hat(r))=-GMmu^(2)=m[(d^(2)r)/((d)t^(2))-r(((d)theta)/((d)t))^(2)]],[(20.32)=-(L^(2))/(m)u^(2)((d^(2)u)/((d)theta^(2))+u)]:}\begin{align*} F_{\hat{r}}=-G M m u^{2} & =m\left[\frac{\mathrm{~d}^{2} r}{\mathrm{~d} t^{2}}-r\left(\frac{\mathrm{~d} \theta}{\mathrm{~d} t}\right)^{2}\right] \\ & =-\frac{L^{2}}{m} u^{2}\left(\frac{\mathrm{~d}^{2} u}{\mathrm{~d} \theta^{2}}+u\right) \tag{20.32} \end{align*}Fr^=GMmu2=m[ d2r dt2r( dθ dt)2](20.32)=L2mu2( d2u dθ2+u)
We end up with a differential equation for the trajectories
(20.33) d 2 u d θ 2 + u = G M m 2 L 2 (20.33) d 2 u d θ 2 + u = G M m 2 L 2 {:(20.33)(d^(2)u)/((d)theta^(2))+u=(GMm^(2))/(L^(2)):}\begin{equation*} \frac{\mathrm{d}^{2} u}{\mathrm{~d} \theta^{2}}+u=\frac{G M m^{2}}{L^{2}} \tag{20.33} \end{equation*}(20.33)d2u dθ2+u=GMm2L2
The solution to this equation is a function u ( θ ) u ( θ ) u(theta)u(\theta)u(θ) that gives the trajectory.
10 10 ^(10){ }^{10}10 The fact that the potential has a minimum implies the motion for particles bound in the potential is stable.
11 11 ^(11){ }^{11}11 The parabola corresponds to the case ϵ = 1 ϵ = 1 epsilon=1\epsilon=1ϵ=1 in eqn 20.11.
Rewriting the differential equation from the last example, we use the variable u 0 = G M m 2 L 2 u 0 = G M m 2 L 2 u_(0)=(GMm^(2))/(L^(2))u_{0}=\frac{G M m^{2}}{L^{2}}u0=GMm2L2, and end up with an equation of motion for the variable u u uuu which is
(20.34) d 2 d θ 2 ( u u 0 ) = ( u u 0 ) (20.34) d 2 d θ 2 u u 0 = u u 0 {:(20.34)(d^(2))/((d)theta^(2))(u-u_(0))=-(u-u_(0)):}\begin{equation*} \frac{\mathrm{d}^{2}}{\mathrm{~d} \theta^{2}}\left(u-u_{0}\right)=-\left(u-u_{0}\right) \tag{20.34} \end{equation*}(20.34)d2 dθ2(uu0)=(uu0)
The general solution of this equation, that gives access to all of the allowed trajectories, is
(20.35) u u 0 = B cos θ (20.35) u u 0 = B cos θ {:(20.35)u-u_(0)=B cos theta:}\begin{equation*} u-u_{0}=B \cos \theta \tag{20.35} \end{equation*}(20.35)uu0=Bcosθ
where B B BBB is a constant.
Now for some interpretation. If we can write the expression for the trajectories in terms of the total energy E E EEE, then we can classify orbits. If the energy is negative we have a bound state or, in other words, an orbit. If the energy is positive, the trajectory is not bounded.
We shall rewrite eqn 20.35 in terms of the values of several physical quantities evaluated at the perihelion (i.e. when θ = 0 θ = 0 theta=0\theta=0θ=0 ). At this point, the radius is r = r 1 r = r 1 r=r_(1)r=r_{1}r=r1 (and so u = u 1 u = u 1 u=u_(1)u=u_{1}u=u1 ) the kinetic energy and potential energy are K 1 K 1 K_(1)K_{1}K1 and U 1 U 1 U_(1)U_{1}U1 respectively and angular momentum is L L LLL. At the perihelion, the radius is perpendicular to the velocity and so the angular momentum is L = m v θ ^ r 1 L = m v θ ^ r 1 L=mv_( hat(theta))r_(1)L=m v_{\hat{\theta}} r_{1}L=mvθ^r1, and so K 1 = L 2 / 2 m K 1 = L 2 / 2 m K_(1)=L^(2)//2mK_{1}=L^{2} / 2 mK1=L2/2m. This allows us to say
(20.36) u 0 = G M m 2 L 2 = U 1 2 K 1 u 1 . (20.36) u 0 = G M m 2 L 2 = U 1 2 K 1 u 1 . {:(20.36)u_(0)=(GMm^(2))/(L^(2))=-(U_(1))/(2K_(1))*u_(1).:}\begin{equation*} u_{0}=\frac{G M m^{2}}{L^{2}}=-\frac{U_{1}}{2 K_{1}} \cdot u_{1} . \tag{20.36} \end{equation*}(20.36)u0=GMm2L2=U12K1u1.
Since when u = u 1 u = u 1 u=u_(1)u=u_{1}u=u1 we have θ = 0 θ = 0 theta=0\theta=0θ=0, eqn 20.35 gives us an expression u 1 u 0 = B u 1 u 0 = B u_(1)-u_(0)=Bu_{1}-u_{0}=Bu1u0=B, which, together with the total energy E = K 1 + U 1 E = K 1 + U 1 E=K_(1)+U_(1)E=K_{1}+U_{1}E=K1+U1, allows us to write and equation linking the constant total energy to the parameters u 0 u 0 u_(0)u_{0}u0 and B B BBB
(20.37) E = u 1 K 1 ( B u 0 ) (20.37) E = u 1 K 1 B u 0 {:(20.37)E=(u_(1))/(K_(1))(B-u_(0)):}\begin{equation*} E=\frac{u_{1}}{K_{1}}\left(B-u_{0}\right) \tag{20.37} \end{equation*}(20.37)E=u1K1(Bu0)
Using this expression, we can classify the three general types of trajectory that are possible, via the sign of E E EEE.
  • When B = u 0 B = u 0 B=u_(0)B=u_{0}B=u0, we have E = 0 E = 0 E=0E=0E=0 and the trajectory is the parabola u = u 0 ( 1 + cos θ ) u = u 0 ( 1 + cos θ ) u=u_(0)(1+cos theta)u=u_{0}(1+\cos \theta)u=u0(1+cosθ). This allows r r rrr to go to infinity when θ = π θ = π theta=pi\theta=\piθ=π so the trajectory is, only just, unbounded. 11 11 ^(11){ }^{11}11
  • When B > u 0 B > u 0 B > u_(0)B>u_{0}B>u0, we have E E EEE positive and the trajectory is a hyperbola. This allows r r rrr to go to infinity when θ = cos 1 ( u 0 / B ) θ = cos 1 u 0 / B theta=cos^(-1)(-u_(0)//B)\theta=\cos ^{-1}\left(-u_{0} / B\right)θ=cos1(u0/B). This is an unbounded trajectory.
  • Finally, in order to have E E EEE negative (resulting in a bounded orbit), we must have B < u 0 B < u 0 B < u_(0)B<u_{0}B<u0. Comparing u u 0 = B cos θ u u 0 = B cos θ u-u_(0)=B cos thetau-u_{0}=B \cos \thetauu0=Bcosθ to eqn 20.11 for the ellipse, we see that it is identical if we take u 0 = a / b 2 u 0 = a / b 2 u_(0)=a//b^(2)u_{0}=a / b^{2}u0=a/b2 and the eccentricity of the ellipse is given by B / u 0 = ϵ B / u 0 = ϵ B//u_(0)=epsilonB / u_{0}=\epsilonB/u0=ϵ. The equation u u 0 = B cos θ u u 0 = B cos θ u-u_(0)=B cos thetau-u_{0}=B \cos \thetauu0=Bcosθ with B < u 0 B < u 0 B < u_(0)B<u_{0}B<u0 then describes an ellipse with limiting radii r = ( u 0 + B ) 1 r = u 0 + B 1 r=(u_(0)+B)^(-1)r=\left(u_{0}+B\right)^{-1}r=(u0+B)1 and r = ( u 0 B ) 1 r = u 0 B 1 r=(u_(0)-B)^(-1)r=\left(u_{0}-B\right)^{-1}r=(u0B)1. (The case of the circular orbit corresponds to B = 0 B = 0 B=0B=0B=0, as it must.)
Example 20.6
Another useful route to the finding the possible trajectories is to start with the equation for the total energy
(20.38) E = 1 2 m r ˙ 2 + 1 2 m r 2 θ ˙ 2 G M m r (20.38) E = 1 2 m r ˙ 2 + 1 2 m r 2 θ ˙ 2 G M m r {:(20.38)E=(1)/(2)mr^(˙)^(2)+(1)/(2)mr^(2)theta^(˙)^(2)-(GMm)/(r):}\begin{equation*} E=\frac{1}{2} m \dot{r}^{2}+\frac{1}{2} m r^{2} \dot{\theta}^{2}-\frac{G M m}{r} \tag{20.38} \end{equation*}(20.38)E=12mr˙2+12mr2θ˙2GMmr
and divide through by L 2 / 2 m L 2 / 2 m L^(2)//2mL^{2} / 2 mL2/2m. Writing u = u / θ u = u / θ u^(')=del u//del thetau^{\prime}=\partial u / \partial \thetau=u/θ we obtain, after a little algebra, that
(20.39) ( u ) 2 + u 2 2 G M m 2 L 2 u 2 E m L 2 = 0 (20.39) u 2 + u 2 2 G M m 2 L 2 u 2 E m L 2 = 0 {:(20.39)(u^('))^(2)+u^(2)-(2GMm^(2))/(L^(2))*u-(2Em)/(L^(2))=0:}\begin{equation*} \left(u^{\prime}\right)^{2}+u^{2}-\frac{2 G M m^{2}}{L^{2}} \cdot u-\frac{2 E m}{L^{2}}=0 \tag{20.39} \end{equation*}(20.39)(u)2+u22GMm2L2u2EmL2=0
Spotting the presence of u 0 = 1 / r 0 = G M m 2 / L 2 u 0 = 1 / r 0 = G M m 2 / L 2 u_(0)=1//r_(0)=GMm^(2)//L^(2)u_{0}=1 / r_{0}=G M m^{2} / L^{2}u0=1/r0=GMm2/L2 we choose to rewrite this quadratic equation as
(20.40) ( u ) 2 + ( u u 0 ) 2 u 0 2 ϵ 2 = 0 (20.40) u 2 + u u 0 2 u 0 2 ϵ 2 = 0 {:(20.40)(u^('))^(2)+(u-u_(0))^(2)-u_(0)^(2)epsilon^(2)=0:}\begin{equation*} \left(u^{\prime}\right)^{2}+\left(u-u_{0}\right)^{2}-u_{0}^{2} \epsilon^{2}=0 \tag{20.40} \end{equation*}(20.40)(u)2+(uu0)2u02ϵ2=0
where u 0 2 ( 1 ϵ 2 ) = 2 E m / L 2 u 0 2 1 ϵ 2 = 2 E m / L 2 u_(0)^(2)(1-epsilon^(2))=-2Em//L^(2)u_{0}^{2}\left(1-\epsilon^{2}\right)=-2 E m / L^{2}u02(1ϵ2)=2Em/L2. Try a solution u = c + d cos θ u = c + d cos θ u=c+d cos thetau=c+d \cos \thetau=c+dcosθ, and find c = u 0 c = u 0 c=u_(0)c=u_{0}c=u0 and d = ± u 0 ϵ d = ± u 0 ϵ d=+-u_(0)epsilond= \pm u_{0} \epsilond=±u0ϵ, so that we have
(20.41) u = u 0 ( 1 ± ϵ cos θ ) (20.41) u = u 0 ( 1 ± ϵ cos θ ) {:(20.41)u=u_(0)(1+-epsilon cos theta):}\begin{equation*} u=u_{0}(1 \pm \epsilon \cos \theta) \tag{20.41} \end{equation*}(20.41)u=u0(1±ϵcosθ)
With our definition of θ θ theta\thetaθ the equation for the ellipse takes the positive sign and we have u = u 0 ( 1 + ϵ cos θ ) u = u 0 ( 1 + ϵ cos θ ) u=u_(0)(1+epsilon cos theta)u=u_{0}(1+\epsilon \cos \theta)u=u0(1+ϵcosθ).

20.5 The why? of orbits

What guarantees that there are orbits at all? That is, why should any trajectory close and hence be periodic? Let's temporarily examine nonrelativistic orbits more generally, taking the central potential energy to be U ( r ) U ( r ) U(r)U(r)U(r), which is not necessarily the Newtonian potential U ( r ) 1 / r U ( r ) 1 / r U(r)prop-1//rU(r) \propto-1 / rU(r)1/r. (In any case, the angular momentum L L vec(L)\vec{L}L is still conserved owing to the lack of torque from a central field.) We can rewrite eqn 20.27 for the total energy as
(20.42) d r d t = { 2 m [ E U ( r ) ] L 2 m 2 r 2 } 1 2 , (20.42) d r d t = 2 m [ E U ( r ) ] L 2 m 2 r 2 1 2 , {:(20.42)(dr)/((d)t)={(2)/(m)[E-U(r)]-(L^(2))/(m^(2)r^(2))}^((1)/(2))",":}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} t}=\left\{\frac{2}{m}[E-U(r)]-\frac{L^{2}}{m^{2} r^{2}}\right\}^{\frac{1}{2}}, \tag{20.42} \end{equation*}(20.42)dr dt={2m[EU(r)]L2m2r2}12,
from which we find the time taken for the motion between two values of r r rrr to be
(20.43) t = r 1 r 2 d r { 2 m [ E U ( r ) ] L 2 m 2 r 2 } 1 2 (20.43) t = r 1 r 2 d r 2 m [ E U ( r ) ] L 2 m 2 r 2 1 2 {:(20.43)t=int_(r_(1))^(r_(2))((d)r)/({(2)/(m)[E-U(r)]-(L^(2))/(m^(2)r^(2))}^((1)/(2))):}\begin{equation*} t=\int_{r_{1}}^{r_{2}} \frac{\mathrm{~d} r}{\left\{\frac{2}{m}[E-U(r)]-\frac{L^{2}}{m^{2} r^{2}}\right\}^{\frac{1}{2}}} \tag{20.43} \end{equation*}(20.43)t=r1r2 dr{2m[EU(r)]L2m2r2}12
or, since d θ = L d t / m r 2 d θ = L d t / m r 2 dtheta=Ldt//mr^(2)\mathrm{d} \theta=L \mathrm{~d} t / m r^{2}dθ=L dt/mr2, we have
(20.44) θ = r 1 r 2 d r L / r 2 { 2 m [ E U ( r ) ] L 2 r 2 } 1 2 (20.44) θ = r 1 r 2 d r L / r 2 2 m [ E U ( r ) ] L 2 r 2 1 2 {:(20.44)theta=int_(r_(1))^(r_(2))((d)rL//r^(2))/({2m[E-U(r)]-(L^(2))/(r^(2))}^((1)/(2))):}\begin{equation*} \theta=\int_{r_{1}}^{r_{2}} \frac{\mathrm{~d} r L / r^{2}}{\left\{2 m[E-U(r)]-\frac{L^{2}}{r^{2}}\right\}^{\frac{1}{2}}} \tag{20.44} \end{equation*}(20.44)θ=r1r2 drL/r2{2m[EU(r)]L2r2}12
If the motion has two limiting radii, r min r min r_(min)r_{\min }rmin and r max r max r_(max)r_{\max }rmax then, during the time in which r r rrr varies from r min r min r_(min)r_{\min }rmin to r max r max r_(max)r_{\max }rmax and back, the radius vector turns through an angle
(20.45) Δ θ = 2 r min r max d r L / r 2 { 2 m [ E U ( r ) ] L 2 r 2 } 1 2 . (20.45) Δ θ = 2 r min r max d r L / r 2 2 m [ E U ( r ) ] L 2 r 2 1 2 . {:(20.45)Delta theta=2int_(r_(min))^(r_(max))(drL//r^(2))/({2m[E-U(r)]-(L^(2))/(r^(2))}^((1)/(2))).:}\begin{equation*} \Delta \theta=2 \int_{r_{\min }}^{r_{\max }} \frac{\mathrm{d} r L / r^{2}}{\left\{2 m[E-U(r)]-\frac{L^{2}}{r^{2}}\right\}^{\frac{1}{2}}} . \tag{20.45} \end{equation*}(20.45)Δθ=2rminrmaxdrL/r2{2m[EU(r)]L2r2}12.
12 12 ^(12){ }^{12}12 The amount of precession per orbit is given by
δ θ = Δ θ 2 π δ θ = Δ θ 2 π delta theta=Delta theta-2pi\delta \theta=\Delta \theta-2 \piδθ=Δθ2π.
Fig. 20.7 The precession of an orbit After n n nnn periods we must make m m mmm complete revolutions in order for the orbit to close.
13 13 ^(13){ }^{13}13 Alternatively, the integral is carried out on page 30 of the book by Zee.
Fig. 20.8 The effective potential for three cases: (a) n < 3 n < 3 n < -3n<-3n<3, (b) 3 < n < 3 < n < -3 < n <-3<n<3<n< -1 , (c) n > 3 n > 3 n > -3n>-3n>3. Stable circular orbits at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 are only possible if n > 3 n > 3 n > -3n>-3n>3.
For this motion to form a closed orbit we must have Δ θ = 2 π q / n Δ θ = 2 π q / n Delta theta=2pi q//n\Delta \theta=2 \pi q / nΔθ=2πq/n with q q qqq and n n nnn integers. That is to say, after n n nnn periods of motion between r min r min  r_("min ")r_{\text {min }}rmin  and r max r max  r_("max ")r_{\text {max }}rmax , the radius vector has made q q qqq complete revolutions in θ θ theta\thetaθ and we're back where we started. This allows for the precession of the orbits, as shown in Fig. 20.7. 12 12 ^(12){ }^{12}12

Example 20.7

We can demonstrate that the orbit closes for the Newtonian case, that is, when U ( r ) = α r U ( r ) = α r U(r)=-(alpha )/(r)U(r)=-\frac{\alpha}{r}U(r)=αr. Rewrite the integral as
(20.47) Δ θ = 2 L r min r max d r { 2 m E + 2 m α r L 2 r 2 } 1 2 (20.47) Δ θ = 2 L r min r max d r 2 m E + 2 m α r L 2 r 2 1 2 {:(20.47)Delta theta=-2(del)/(del L)int_(r_(min))^(r_(max))dr{2mE+(2m alpha)/(r)-(L^(2))/(r^(2))}^((1)/(2)):}\begin{equation*} \Delta \theta=-2 \frac{\partial}{\partial L} \int_{r_{\min }}^{r_{\max }} \mathrm{d} r\left\{2 m E+\frac{2 m \alpha}{r}-\frac{L^{2}}{r^{2}}\right\}^{\frac{1}{2}} \tag{20.47} \end{equation*}(20.47)Δθ=2Lrminrmaxdr{2mE+2mαrL2r2}12
This can be rewritten as an elliptical integral whose result can be looked up: 13 13 ^(13){ }^{13}13 it turns out to yield Δ θ ( = 2 π q / n ) = 2 π Δ θ ( = 2 π q / n ) = 2 π Delta theta(=2pi q//n)=2pi\Delta \theta(=2 \pi q / n)=2 \piΔθ(=2πq/n)=2π. This implies that q = n q = n q=nq=nq=n and there is no precession of any Newtonian orbit.
Another way of looking at the closing of trajectories is covered in the following example.

Example 20.8

For a central field of force F = A r n F = A r n F=-Ar^(n)F=-A r^{n}F=Arn, the effective potential is
(20.48) V eff = A r n + 1 n + 1 + L 2 2 m r 2 (20.48) V eff = A r n + 1 n + 1 + L 2 2 m r 2 {:(20.48)V_(eff)=(Ar^(n+1))/(n+1)+(L^(2))/(2mr^(2)):}\begin{equation*} V_{\mathrm{eff}}=\frac{A r^{n+1}}{n+1}+\frac{L^{2}}{2 m r^{2}} \tag{20.48} \end{equation*}(20.48)Veff=Arn+1n+1+L22mr2
This gives a stable circular orbit at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 only if n > 3 n > 3 n > -3n>-3n>3. (The form of the effective potential is shown in Fig. 20.8 for three distinct cases.) Near a circular orbit at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0, we can write the effective potential as a Taylor expansion
(20.49) V eff = V ( r 0 ) + 1 2 ( r r 0 ) 2 ( d 2 V eff d r 2 ) r = r 0 + (20.49) V eff = V r 0 + 1 2 r r 0 2 d 2 V eff d r 2 r = r 0 + {:(20.49)V_(eff)=V(r_(0))+(1)/(2)(r-r_(0))^(2)((d^(2)V_(eff))/((d)r^(2)))_(r=r_(0))+cdots:}\begin{equation*} V_{\mathrm{eff}}=V\left(r_{0}\right)+\frac{1}{2}\left(r-r_{0}\right)^{2}\left(\frac{\mathrm{~d}^{2} V_{\mathrm{eff}}}{\mathrm{~d} r^{2}}\right)_{r=r_{0}}+\cdots \tag{20.49} \end{equation*}(20.49)Veff=V(r0)+12(rr0)2( d2Veff dr2)r=r0+
where d 2 V eff / d r 2 d 2 V eff  / d r 2 d^(2)V_("eff ")//dr^(2)\mathrm{d}^{2} V_{\text {eff }} / \mathrm{d} r^{2}d2Veff /dr2 evaluated at r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0 is given by ( n + 3 ) L 2 / ( m r 0 4 ) ( n + 3 ) L 2 / m r 0 4 (n+3)L^(2)//(mr_(0)^(4))(n+3) L^{2} /\left(m r_{0}^{4}\right)(n+3)L2/(mr04). Small oscillations for a particle of mass m m mmm near the bottom of the well, i.e. orbiting close to r = r 0 r = r 0 r=r_(0)r=r_{0}r=r0, are therefore governed by
(20.50) m r ¨ + ( n + 3 ) L 2 m r 0 4 ( r r 0 ) = 0 (20.50) m r ¨ + ( n + 3 ) L 2 m r 0 4 r r 0 = 0 {:(20.50)mr^(¨)+((n+3)L^(2))/(mr_(0)^(4))*(r-r_(0))=0:}\begin{equation*} m \ddot{r}+\frac{(n+3) L^{2}}{m r_{0}^{4}} \cdot\left(r-r_{0}\right)=0 \tag{20.50} \end{equation*}(20.50)mr¨+(n+3)L2mr04(rr0)=0
which is simple harmonic motion with period τ τ tau\tauτ given by
(20.51) τ = 2 π m r 0 2 ( n + 3 ) 1 / 2 L (20.51) τ = 2 π m r 0 2 ( n + 3 ) 1 / 2 L {:(20.51)tau=(2pi mr_(0)^(2))/((n+3)^(1//2)L):}\begin{equation*} \tau=\frac{2 \pi m r_{0}^{2}}{(n+3)^{1 / 2} L} \tag{20.51} \end{equation*}(20.51)τ=2πmr02(n+3)1/2L
The orbital period T = 2 π / ω = 2 π m r 0 2 / L T = 2 π / ω = 2 π m r 0 2 / L T=2pi//omega=2pi mr_(0)^(2)//LT=2 \pi / \omega=2 \pi m r_{0}^{2} / LT=2π/ω=2πmr02/L and so these two periods are related by
(20.52) τ = T ( n + 3 ) 1 / 2 . (20.52) τ = T ( n + 3 ) 1 / 2 . {:(20.52)tau=(T)/((n+3)^(1//2)).:}\begin{equation*} \tau=\frac{T}{(n+3)^{1 / 2}} . \tag{20.52} \end{equation*}(20.52)τ=T(n+3)1/2.
  • For the n = 2 n = 2 n=-2n=-2n=2 (Newtonian) case, τ = T τ = T tau=T\tau=Tτ=T and so the precession leads to elliptical orbits. Another stable case occurs for n = 1 n = 1 n=1n=1n=1 (simple harmonic motion) where τ = T / 2 τ = T / 2 tau=T//2\tau=T / 2τ=T/2.
  • For general n > 3 n > 3 n > -3n>-3n>3, excepting these special cases, the precession τ τ tau\tauτ is incommensurate with the orbital motion. For this reason, any departures from n = 2 n = 2 n=-2n=-2n=2 in some imagined small departure from Newtonian gravity should be easy to determine from observations since it will lead to precession of elliptical orbits.
The closing of trajectories is further demonstrated for Newtonian orbits in the next example.

Example 20.9

We start by proving a geometric result. The Laplace-Runge-Lenz vector 14 14 ^(14){ }^{14}14 is defined as
(20.53) L = ( p × L ) G M m 2 r r (20.53) L = ( p × L ) G M m 2 r r {:(20.53) vec(L)=( vec(p)xx vec(L))-GMm^(2)(( vec(r)))/(r):}\begin{equation*} \overrightarrow{\mathcal{L}}=(\vec{p} \times \vec{L})-G M m^{2} \frac{\vec{r}}{r} \tag{20.53} \end{equation*}(20.53)L=(p×L)GMm2rr
Taking the time derivative of this vector, we find
L ˙ = ( F × L ) + ( p × L ˙ ) G M m 2 r ˙ r + G M m 2 r r 2 r ˙ (time derivative) = F × r × p G M m 2 r ˙ r + G M m 2 r r 2 r ˙ (using τ = L ˙ = 0 ) = r ( F p ) p ( F r ) G M m 2 r r + G M m 2 r r 2 r ˙ (triple vector prod = G M m 2 r r 2 r ˙ + G M m 2 r r G M m 2 r r + G M m 2 r r 2 r ˙ = 0 . L ˙ = ( F × L ) + ( p × L ˙ ) G M m 2 r ˙ r + G M m 2 r r 2 r ˙  (time derivative)  = F × r × p G M m 2 r ˙ r + G M m 2 r r 2 r ˙  (using  τ = L ˙ = 0  )  = r ( F p ) p ( F r ) G M m 2 r r + G M m 2 r r 2 r ˙  (triple vector prod  = G M m 2 r r 2 r ˙ + G M m 2 r r G M m 2 r r + G M m 2 r r 2 r ˙ = 0 . {:[ vec(L)^(˙)=( vec(F)xx vec(L))+( vec(p)xx vec(L)^(˙))-GMm^(2)(( vec(r)^(˙)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)" (time derivative) "],[= vec(F)xx vec(r)xx vec(p)-GMm^(2)((r^(˙)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)" (using "tau= vec(L)^(˙)=0" ) "],[= vec(r)( vec(F)* vec(p))- vec(p)( vec(F)* vec(r))-GMm^(2)(( vec(r)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)" (triple vector prod "],[=-GMm^(2)(( vec(r)))/(r^(2))r^(˙)+GMm^(2)(( vec(r)))/(r)-GMm^(2)(( vec(r)))/(r)+GMm^(2)(( vec(r)))/(r^(2))r^(˙)],[=0.]:}\begin{aligned} \dot{\overrightarrow{\mathcal{L}}} & =(\vec{F} \times \vec{L})+(\vec{p} \times \dot{\vec{L}})-G M m^{2} \frac{\dot{\vec{r}}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \text { (time derivative) } \\ & =\vec{F} \times \vec{r} \times \vec{p}-G M m^{2} \frac{\dot{r}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \text { (using } \tau=\dot{\vec{L}}=0 \text { ) } \\ & =\vec{r}(\vec{F} \cdot \vec{p})-\vec{p}(\vec{F} \cdot \vec{r})-G M m^{2} \frac{\vec{r}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \text { (triple vector prod } \\ & =-G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r}+G M m^{2} \frac{\vec{r}}{r}-G M m^{2} \frac{\vec{r}}{r}+G M m^{2} \frac{\vec{r}}{r^{2}} \dot{r} & & \\ & =0 . & & \end{aligned}L˙=(F×L)+(p×L˙)GMm2r˙r+GMm2rr2r˙ (time derivative) =F×r×pGMm2r˙r+GMm2rr2r˙ (using τ=L˙=0 ) =r(Fp)p(Fr)GMm2rr+GMm2rr2r˙ (triple vector prod =GMm2rr2r˙+GMm2rrGMm2rr+GMm2rr2r˙=0.
Now consider the orbit at the perihelion (and aphelion). Here we have that the momentum p p vec(p)\vec{p}p is perpendicular to the semi-major axis of the ellipse. Since L L vec(L)\vec{L}L points out of the plane of the orbit we can take L L vec(L)\overrightarrow{\mathcal{L}}L to point from the focus to the perihelion. Since the vector L L vec(L)\overrightarrow{\mathcal{L}}L is constant, it always points to the perihelion and so the perihelion cannot move. The orbit never precesses and must therefore close. 15 15 ^(15){ }^{15}15
With this review of Newtonian orbits under our belts, we can examine some of the richness afforded by general relativity's correction to the Newtonian picture. In the next chapter, we meet a metric field analogous to the spherically symmetric 1 / r 1 / r 1//r1 / r1/r potential of the Newtonian case. This is the celebrated Schwarzschild metric.

Chapter summary

  • Orbits in a Newtonian potential are elliptical. They can be understood by identifying an effective potential.
  • The variable u = 1 / r u = 1 / r u=1//ru=1 / ru=1/r allows us to rewrite the problem and solve the equations of motion. We make use of conserved energy and angular momentum.
  • Newtonian orbits never precess.
    14 14 ^(14){ }^{14}14 Pierre-Simon Laplace (1749-1827), Carl Runge (1856-1927) and Wilhelm Lenz (1888-1957). The vector is also known as the Laplace vector, the Runge-Lenz vector and the Lenz vector. In fact, the quantity seems to predate all of these people. Herbert Goldstein's two short articles [Am. J. Phys, 43, 737 (1975) and 44, 1123 (1976)] 43, 737 (1975) and 44, 1123 (1976)] outline its interesting history, where it is suggested that priority actually beongs to Jakob Hermann (1678-1733) and Johann Bernoulli (1667-1748). As discussed in the book by Gutzwiller Wolfgang Pauli used the properties of L L vec(L)\overrightarrow{\mathcal{L}}L to compute the quantum-mechanical spectrum of the hydrogen atom exactly in 1926.
    15 15 ^(15){ }^{15}15 The components of the vector L L vec(L)\overrightarrow{\mathcal{L}}L, along with the energy E E EEE, and the components of the angular momentum L L vec(L)\vec{L}L give us seven quantities. In the Kepler give us seven quantities. vec(vec())\overrightarrow{\vec{~}}  the Kepler problem, the length of L L L\mathcal{L}L is constant and we also have L L ~ = 0 L L ~ = 0 L* tilde(L)=0\mathcal{L} \cdot \tilde{L}=0LL~=0, giving us five independent constants of the motion. In general, a mechanical systems with d d ddd degrees of freedom can have, at most, 2 d 1 2 d 1 2d-12 d-12d1 constants of the motion. [This is because there are 2 d 2 d 2d2 d2d initial conditions (the components of position and velocity) and the initial time cannot be determined by a constant of the motion.] The Kepler problem has d = 3 d = 3 d=3d=3d=3, so we have the maximum number of constants of the motion possible. For this reason, the Kepler problem is sometimes called 'maximally superintegrable'.

Exercises

(20.1) Fill in the algebra leading to eqn 20.39 .
(20.2) Two particles, each of mass m m mmm, move under the influence of their mutual gravitational attraction
G m 2 r 2 G m 2 r 2 -(Gm^(2))/(r^(2))-\frac{G m^{2}}{r^{2}}Gm2r2. Initially, the particles are a large distance apart and approach each other with velocities v v vec(v)\vec{v}v and v v - vec(v)-\vec{v}v along parallel paths a distance b b bbb apart.
(a) What is the angular momentum of the system during the motion?
(b) Write down the energy of the system at the start of the motion and at the point in the motion when the particles are closest to each other.
(c) Show that the least distance d d ddd between the particles in their subsequent motion is given by
d = G m 2 v 2 + ( G m 2 v 2 ) 2 + b 2 d = G m 2 v 2 + G m 2 v 2 2 + b 2 d=-(Gm)/(2v^(2))+sqrt(((Gm)/(2v^(2)))^(2)+b^(2))d=-\frac{G m}{2 v^{2}}+\sqrt{\left(\frac{G m}{2 v^{2}}\right)^{2}+b^{2}}d=Gm2v2+(Gm2v2)2+b2
(20.3) A particle with mass m m mmm is placed a distance x x xxx from the centre of a thin ring of radius a a aaa, along the line through the centre of the ring and perpendicular to its plane.
(a) Assuming the ring has a total mass M M MMM, which is uniformly distributed along the ring, show that the total gravitational potential energy of the ring and particle is given by
(20.54) U = G M m ( a 2 + x 2 ) 1 2 (20.54) U = G M m a 2 + x 2 1 2 {:(20.54)U=-(GMm)/((a^(2)+x^(2))^((1)/(2))):}\begin{equation*} U=-\frac{G M m}{\left(a^{2}+x^{2}\right)^{\frac{1}{2}}} \tag{20.54} \end{equation*}(20.54)U=GMm(a2+x2)12
(b) Find the magnitude and direction of the force on the particle. Comment on this result in the case that the distance between the particle and ring is very large compared to the ring's radius.
Now consider a mass Ω Ω Omega\OmegaΩ uniformly distributed over a disc of radius L L LLL. A particle of mass m m mmm is placed a distance x x xxx from the centre of the disc, along the line through its centre and perpendicular to its plane.
(c) Using the result from part (a), or otherwise,
find the gravitational potential energy of the disc and particle system.
(20.4) (a) Show that an equation of motion
(20.55) d p α d τ = m Φ x α (20.55) d p α d τ = m Φ x α {:(20.55)(dp_(alpha))/(dtau)=-m(del Phi)/(delx^(alpha)):}\begin{equation*} \frac{\mathrm{d} p_{\alpha}}{\mathrm{d} \tau}=-m \frac{\partial \Phi}{\partial x^{\alpha}} \tag{20.55} \end{equation*}(20.55)dpαdτ=mΦxα
where Φ Φ Phi\PhiΦ is the time-independent Newtonian potential, is incompatible with the rule from special relativity that a u = 0 a u = 0 a*u=0\boldsymbol{a} \cdot \boldsymbol{u}=0au=0 (where a a a\boldsymbol{a}a is acceleration and u u u\boldsymbol{u}u is velocity).
(b) Show that the equation of motion
(20.56) d p α d τ = m ( η α β + u α u β ) Φ x β (20.56) d p α d τ = m η α β + u α u β Φ x β {:(20.56)(dp^(alpha))/(dtau)=-m(eta^(alpha beta)+u^(alpha)u^(beta))(del Phi)/(delx^(beta)):}\begin{equation*} \frac{\mathrm{d} p^{\alpha}}{\mathrm{d} \tau}=-m\left(\eta^{\alpha \beta}+u^{\alpha} u^{\beta}\right) \frac{\partial \Phi}{\partial x^{\beta}} \tag{20.56} \end{equation*}(20.56)dpαdτ=m(ηαβ+uαuβ)Φxβ
does not suffer from the problem in (a).
The latter can be regarded as an alternative theory of gravity in which the gravitational field exists in Minkowski spacetime. The projection operator ( η + u u ) ( η + u u ) (eta+u ox u)(\boldsymbol{\eta}+\boldsymbol{u} \otimes \boldsymbol{u})(η+uu) picks out the part perpendicular to u u u\boldsymbol{u}u (i.e. perpendicular to the world line, ensuring the orthogonality of this part and u u u\boldsymbol{u}u ).
(20.5) Consider the equation of motion from Exercise 20.4(b). By taking the limit of zero mass, such that λ = τ / m λ = τ / m lambda=tau//m\lambda=\tau / mλ=τ/m remains constant as m m mmm and τ τ tau\tauτ go to zero, show that p α e Φ ( r ) p α e Φ ( r ) p^(alpha)e^(Phi(r))p^{\alpha} \mathrm{e}^{\Phi(r)}pαeΦ(r) (for α = 0 4 α = 0 4 alpha=0-4\alpha=0-4α=04 ) remains constant along the world line of a photon. This approach (based on that of Thorne and Blandford) shows that light cannot be deflected by the Sun in this theory, in contradiction to experiment. Gravity cannot, therefore, simply be incorporated into special relativity as another field in Minkowski spacetime.

The Schwarzschild geometry

As you see the war treated me kindly enough, in spite of heavy gunfire, to allow me to get away from it all and take this walk in the land of your ideas.
Karl Schwarzschild (1873-1916) in a 1915 letter to Einstein
After Einstein wrote down the field equation of general relativity he did not expect it to admit exact solutions owing to its complexity. He himself used an approximate solution in his 1915 article about the perihelion of Mercury. It therefore came as something of a surprise when Einstein received a letter from Karl Schwarzschild at the end of 1915 detailing a rather simple exact solution. 1 1 ^(1){ }^{1}1 Schwarzschild was, despite being over forty years old, then serving as a soldier in World War I and carried out his work while at the Russian front.
The Schwarzschild solution is a solution of the Einstein field equation for the geometry outside a spherically symmetric, gravitating mass distribution. (We will find that the solution inside a mass distribution is different.) The Schwarzschild solution gives us a useful metric tensor that we will use to describe the spacetime outside stars and black holes, along with the motion of objects orbiting these bodies. First, the answer: the Schwarzschild metric line element for the space outside a static, spherically symmetric gravitating body of mass M M MMM at the origin of a set of coordinates ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ) is written as 2 2 ^(2){ }^{2}2
(21.1) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.1) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.1)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.1} \end{equation*}(21.1)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
Notable features of this line element include: (i) it is static: none of the components 3 g μ ν 3 g μ ν ^(3)g_(mu nu){ }^{3} g_{\mu \nu}3gμν of the metric field depend on time; (ii) it is asymptotically flat: as r r r rarr oor \rightarrow \inftyr it looks like the Minkowski metric; and (iii) it appears badly behaved at the origin, and also when r = 2 M r = 2 M r=2Mr=2 Mr=2M, where the second term becomes singular. One additional reason why this line element is so special is that Birkhoff's theorem 4 4 ^(4){ }^{4}4 tells us that any spherically symmetric solution to the Einstein equation outside a gravitating object (they don't need to be static, for example) will be identical to Schwarzschild's static solution.
In this chapter, we will examine where the Schwarzschild solution comes from, and how we justify its form. Our results will be useful in the later chapters in this part of the book, where we will apply the
21.1 Justifying the solution 230 21.2 Components of the Riemann tensor 231 21.3 A gravitating object 232 21.4 The meaning of the coordinates

Chapter summary 235

Exercises 235
1 1 ^(1){ }^{1}1 In 1915, Schwarzschild started suffering from pemphigus, a rare autoimmune skin disease that likely led to his death in 1916. English speakers should consider saying 'shvarts-shilt' rather consider saying shvarts-shilt rather
than the commonly heard 'shwortschild'. The name means black shield rather than black child, in any case.
2 2 ^(2){ }^{2}2 Our choice of units in this part of the book is G = c = 1 G = c = 1 G=c=1G=c=1G=c=1. To obtain real-world units in the metric substitute M G M / c 2 M G M / c 2 M rarr GM//c^(2)M \rightarrow G M / c^{2}MGM/c2 and t c t t c t t rarr ctt \rightarrow c ttct.
3 3 ^(3){ }^{3}3 The non-zero components of the metric tensor are
g t t = ( 1 2 M r ) g r r = ( 1 2 M r ) 1 g θ θ = r 2 g ϕ ϕ = r 2 sin 2 θ g t t = 1 2 M r g r r = 1 2 M r 1 g θ θ = r 2 g ϕ ϕ = r 2 sin 2 θ {:[g_(tt)=-(1-(2M)/(r))],[g_(rr)=(1-(2M)/(r))^(-1)],[g_(theta theta)=r^(2)],[g_(phi phi)=r^(2)sin^(2)theta]:}\begin{aligned} & g_{t t}=-\left(1-\frac{2 M}{r}\right) \\ & g_{r r}=\left(1-\frac{2 M}{r}\right)^{-1} \\ & g_{\theta \theta}=r^{2} \\ & g_{\phi \phi}=r^{2} \sin ^{2} \theta \end{aligned}gtt=(12Mr)grr=(12Mr)1gθθ=r2gϕϕ=r2sin2θ
4 4 ^(4){ }^{4}4 George David Birkhoff (1884-1944). Birkhoff's theorem says that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat.
5 5 ^(5){ }^{5}5 We'll postpone examining the light cone structure of this metric field until then.
In non-natural units, the Schwarzschild metric is given by
d s 2 = ( 1 2 G M c 2 r ) c 2 d t 2 + ( 1 2 G M c 2 r ) 1 d r 2 (21.2) + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 G M c 2 r c 2 d t 2 + 1 2 G M c 2 r 1 d r 2 (21.2) + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-(1-(2GM)/(c^(2)r))c^(2)dt^(2)],[+(1-(2GM)/(c^(2)r))^(-1)dr^(2)],[(21.2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{align*} \mathrm{d} s^{2}= & -\left(1-\frac{2 G M}{c^{2} r}\right) c^{2} \mathrm{~d} t^{2} \\ & +\left(1-\frac{2 G M}{c^{2} r}\right)^{-1} \mathrm{~d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.2} \end{align*}ds2=(12GMc2r)c2 dt2+(12GMc2r)1 dr2(21.2)+r2( dθ2+sin2θ dϕ2)
Note that, unlike the coordinate systems we met in the previous part of the book, Schwarzschild coordinates are book, Schore since 0 0 !=0\neq 00 ar not comoving, since g t t , r 0 g t t , r 0 g_(tt,r)!=0g_{t t, r} \neq 0gtt,r0. Physi cally, this is because the Schwarzschild coordinates privilege the rest frame of the spherically symmetric mass distribution that acts as the source of the curvature.
6 6 ^(6){ }^{6}6 Recall our slogan that coordinate have no intrinsic significance in the metric field, so one radius-like variable is just as good as another.
7 7 ^(7){ }^{7}7 This also provides us with an opera tional definition of when a field can be characterized as being weak (i.e. restorcharacterized as being weak (i.e. re
ing factors, we want 2 Φ / c 2 1 2 Φ / c 2 1 2Phi//c^(2)≪12 \Phi / c^{2} \ll 12Φ/c21 ).
metric and use the curvature tensor to examine motion such as orbits and objects like black holes. 5 5 ^(5){ }^{5}5

21.1 Justifying the solution

We seek a solution to the Einstein equation for a static, spherically symmetric mass distribution centred on the origin of a set of spherical coordinates ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ). Although, as we have discussed, the names of these components have no intrinsic metric significance, it will be helpful to give them some temporarily in an effort to make sensible assumptions. We start by assuming that the metric field g g g\boldsymbol{g}g is static. This means that intervals between events are time independent and so components obey d g α β / d t = 0 d g α β / d t = 0 dg_(alpha beta)//dt=0\mathrm{d} g_{\alpha \beta} / \mathrm{d} t=0dgαβ/dt=0. We assume that the spherical symmetry means that world lines of constant r , θ r , θ r,thetar, \thetar,θ and ϕ ϕ phi\phiϕ are orthogonal to t = t = t=t=t= (const.) hypersurfaces. We also assume asymptotic flatness, which is to say that as r r r rarr oor \rightarrow \inftyr, we must have flat spacetime (that is, the gravitational forces vanish at infinity).
A good place to start is therefore with spherically symmetric, flat spacetime, which obeys the above assumptions and has a metric line element
(21.3) d s 2 = d t 2 + d r 2 + r 2 d Ω 2 (21.3) d s 2 = d t 2 + d r 2 + r 2 d Ω 2 {:(21.3)ds^(2)=-dt^(2)+dr^(2)+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} t^{2}+\mathrm{d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{21.3} \end{equation*}(21.3)ds2=dt2+dr2+r2 dΩ2
where d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}dΩ2=dθ2+sin2θ dϕ2. The simplest way to proceed is then to guess that these components assume different values close to the gravitating object, subject to obeying the constraints of the previous paragraph. Owing to spherical symmetry, the new components can only depend on the radius coordinate r r rrr and so we write an expression with three functions
(21.4) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + R ( r ) 2 d Ω 2 (21.4) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + R ( r ) 2 d Ω 2 {:(21.4)ds^(2)=-e^(2Phi(r))dt^(2)+e^(2Lambda(r))dr^(2)+R(r)^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{2 \Phi(r)} \mathrm{d} t^{2}+\mathrm{e}^{2 \Lambda(r)} \mathrm{d} r^{2}+R(r)^{2} \mathrm{~d} \Omega^{2} \tag{21.4} \end{equation*}(21.4)ds2=e2Φ(r)dt2+e2Λ(r)dr2+R(r)2 dΩ2
We can immediately now restrict the number of variables from three to two. This is because we can simply reinterpret R ( r ) R ( r ) R(r)R(r)R(r) as the r r rrr coordinate and rescale the other functions. 6 6 ^(6){ }^{6}6 So we have a general line element
(21.5) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + r 2 d Ω 2 (21.5) d s 2 = e 2 Φ ( r ) d t 2 + e 2 Λ ( r ) d r 2 + r 2 d Ω 2 {:(21.5)ds^(2)=-e^(2Phi(r))dt^(2)+e^(2Lambda(r))dr^(2)+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{2 \Phi(r)} \mathrm{d} t^{2}+\mathrm{e}^{2 \Lambda(r)} \mathrm{d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{21.5} \end{equation*}(21.5)ds2=e2Φ(r)dt2+e2Λ(r)dr2+r2 dΩ2
where we constrain Φ ( ) = Λ ( ) = 0 Φ ( ) = Λ ( ) = 0 Phi(oo)=Lambda(oo)=0\Phi(\infty)=\Lambda(\infty)=0Φ()=Λ()=0.
Expressions for Φ ( r ) Φ ( r ) Phi(r)\Phi(r)Φ(r) and Λ ( r ) Λ ( r ) Lambda(r)\Lambda(r)Λ(r) will come from linking the line element in eqn 21.5 to the physics of gravitating objects, which we do below. However, we can immediately gain an insight into the function Φ ( r ) Φ ( r ) Phi(r)\Phi(r)Φ(r) by comparison with the weak-field metric, which has a line element, expressed in spherical coordinates, of
(21.6) d s 2 = ( 1 + 2 Φ ) d t 2 + d r 2 + r 2 d Ω 2 (21.6) d s 2 = ( 1 + 2 Φ ) d t 2 + d r 2 + r 2 d Ω 2 {:(21.6)ds^(2)=-(1+2Phi)dt^(2)+dr^(2)+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-(1+2 \Phi) \mathrm{d} t^{2}+\mathrm{d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{21.6} \end{equation*}(21.6)ds2=(1+2Φ)dt2+dr2+r2 dΩ2
where Φ Φ Phi\PhiΦ is the Newtonian gravitational potential. In eqn 21.5, we observe that in the limit of small Φ Φ Phi\PhiΦ we have that e 2 Φ ( 1 + 2 Φ ) e 2 Φ ( 1 + 2 Φ ) -e^(2Phi)~~-(1+2Phi)-\mathrm{e}^{2 \Phi} \approx-(1+2 \Phi)e2Φ(1+2Φ), suggesting that this is the same gravitational potential Φ Φ Phi\PhiΦ that features in the weak-field metric. 7 7 ^(7){ }^{7}7
Now that we have, in eqn 21.5, a candidate metric, we can feed it through the Einstein field equation G = 8 π T G = 8 π T G=8pi T\boldsymbol{G}=8 \pi \boldsymbol{T}G=8πT. This allows us to discover that the metric does indeed solve the problem and also how to provide expressions for Φ Φ Phi\PhiΦ and Λ Λ Lambda\LambdaΛ in terms of physically meaningful quantities.

21.2 Components of the Riemann tensor

We start by finding the left-hand side of the Einstein field equation. This involves finding the curvature tensor R R R\boldsymbol{R}R for the metric field in eqn 21.5 and then generating the Einstein tensor G G G\boldsymbol{G}G. The components of R R R\boldsymbol{R}R can be generated in a number of ways. To maximize simplicity, we will work in the orthonormal frame. 8 8 ^(8){ }^{8}8
Example 21.1
Written in its general form, the spherically symmetric metric line element looks like
(21.7) d s 2 = e 2 Φ d t 2 + e 2 Λ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.7) d s 2 = e 2 Φ d t 2 + e 2 Λ d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.7)ds^(2)=-e^(2Phi)dt^(2)+e^(2Lambda)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{2 \Phi} \mathrm{~d} t^{2}+\mathrm{e}^{2 \Lambda} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.7} \end{equation*}(21.7)ds2=e2Φ dt2+e2Λ dr2+r2( dθ2+sin2θ dϕ2)
where Φ Φ Phi\PhiΦ and Λ Λ Lambda\LambdaΛ are functions of r r rrr only. Feeding this into the equations to find the components of the Riemann tensor, we obtain 9 9 ^(9){ }^{9}9
R t r ^ t ^ r ^ = E , R t θ ^ i θ ^ θ ^ = E ¯ , R i t ^ t ^ ϕ ^ = E ¯ , (21.8) R θ ^ ϕ ^ θ ^ ϕ ^ = F , R r ^ ^ ϕ ^ ϕ ^ r ^ = F ¯ , R r ^ θ ^ r ^ θ ^ ^ = F ¯ , R t r ^ t ^ r ^ = E , R t θ ^ i θ ^ θ ^ = E ¯ , R i t ^ t ^ ϕ ^ = E ¯ , (21.8) R θ ^ ϕ ^ θ ^ ϕ ^ = F , R r ^ ^ ϕ ^ ϕ ^ r ^ = F ¯ , R r ^ θ ^ r ^ θ ^ ^ = F ¯ , {:[R^(t hat(r))_( hat(t) hat(r))=E","quadR^(t hat(theta))_(i hat(theta) hat(theta))= bar(E)","quadR^(i hat(t))_( hat(t) hat(phi))= bar(E)","],[(21.8)R^( hat(theta) hat(phi))_( hat(theta) hat(phi))=F","quadR_( hat(hat(r)) hat(phi))^( hat(phi) hat(r))=- bar(F)","quadR^( hat(r) hat(theta)) hat(( hat(r))( hat(theta)))= bar(F)","]:}\begin{align*} & R^{t \hat{r}}{ }_{\hat{t} \hat{r}}=E, \quad R^{t \hat{\theta}}{ }_{i \hat{\theta} \hat{\theta}}=\bar{E}, \quad R^{i \hat{t}}{ }_{\hat{t} \hat{\phi}}=\bar{E}, \\ & R^{\hat{\theta} \hat{\phi}}{ }_{\hat{\theta} \hat{\phi}}=F, \quad R_{\hat{\hat{r}} \hat{\phi}}^{\hat{\phi} \hat{r}}=-\bar{F}, \quad R^{\hat{r} \hat{\theta}} \hat{\hat{r} \hat{\theta}}=\bar{F}, \tag{21.8} \end{align*}Rtr^t^r^=E,Rtθ^iθ^θ^=E¯,Rit^t^ϕ^=E¯,(21.8)Rθ^ϕ^θ^ϕ^=F,Rr^^ϕ^ϕ^r^=F¯,Rr^θ^r^θ^^=F¯,
where
E = e 2 Λ ( Φ + Φ 2 Φ Λ ) E ¯ = 1 r e 2 Λ Φ F = 1 r 2 ( 1 e 2 Λ ) (21.9) F ¯ = 1 r e 2 Λ Λ E = e 2 Λ Φ + Φ 2 Φ Λ E ¯ = 1 r e 2 Λ Φ F = 1 r 2 1 e 2 Λ (21.9) F ¯ = 1 r e 2 Λ Λ {:[E=-e^(-2Lambda)(Phi^('')+Phi^('2)-Phi^(')Lambda^('))],[ bar(E)=-(1)/(r)e^(-2Lambda)Phi^(')],[F=(1)/(r^(2))(1-e^(-2Lambda))],[(21.9) bar(F)=(1)/(r)e^(-2Lambda)Lambda^(')]:}\begin{align*} E & =-\mathrm{e}^{-2 \Lambda}\left(\Phi^{\prime \prime}+\Phi^{\prime 2}-\Phi^{\prime} \Lambda^{\prime}\right) \\ \bar{E} & =-\frac{1}{r} \mathrm{e}^{-2 \Lambda} \Phi^{\prime} \\ F & =\frac{1}{r^{2}}\left(1-\mathrm{e}^{-2 \Lambda}\right) \\ \bar{F} & =\frac{1}{r} \mathrm{e}^{-2 \Lambda} \Lambda^{\prime} \tag{21.9} \end{align*}E=e2Λ(Φ+Φ2ΦΛ)E¯=1re2ΛΦF=1r2(1e2Λ)(21.9)F¯=1re2ΛΛ
and we have used the dash notation for derivatives with respect to r r rrr.
These components of the Riemann tensor R R R\boldsymbol{R}R give us the non-zero components of the Einstein tensor 10 10 ^(10){ }^{10}10
G i t t ^ G i t (21.10) = ( F + 2 F ¯ ) G r ^ r ^ = G r ^ r ^ = ( F + 2 E ¯ ) G θ ^ θ ^ = G ϕ ˙ ϕ ^ = ( E + E ¯ + F ¯ ) G i t t ^ G i t (21.10) = ( F + 2 F ¯ ) G r ^ r ^ = G r ^ r ^ = ( F + 2 E ¯ ) G θ ^ θ ^ = G ϕ ˙ ϕ ^ = ( E + E ¯ + F ¯ ) {:[G_(i)^(t)],[_( hat(t))-G_(it)],[(21.10)=-(F+2 bar(F))],[G_( hat(r))^( hat(r))=G_( hat(r) hat(r))],[=-(F+2 bar(E))],[G_( hat(theta))^( hat(theta))=G_(phi^(˙))^( hat(phi))=-(E+ bar(E)+ bar(F))]:}\begin{align*} & G_{i}^{t} \\ &{ }_{\hat{t}}-G_{i t} \\ &=-(F+2 \bar{F}) \tag{21.10}\\ & G_{\hat{r}}^{\hat{r}}=G_{\hat{r} \hat{r}} \\ &=-(F+2 \bar{E}) \\ & G_{\hat{\theta}}^{\hat{\theta}}=G_{\dot{\phi}}^{\hat{\phi}}=-(E+\bar{E}+\bar{F}) \end{align*}Gitt^Git(21.10)=(F+2F¯)Gr^r^=Gr^r^=(F+2E¯)Gθ^θ^=Gϕ˙ϕ^=(E+E¯+F¯)
Note that G G G\boldsymbol{G}G will vanish outside a mass distribution, although we can have curvature (i.e. non-zero R R R\boldsymbol{R}R ) this region. The tensor G G G\boldsymbol{G}G is zero here because the Einstein equation forces G G G\boldsymbol{G}G to be proportional to the energy-momentum tensor T T T\boldsymbol{T}T, which itself vanishes in the vacuum.
8 8 ^(8){ }^{8}8 Remember that one advantage of this is that components can be raised and lowered with the Minkowski tensor η η eta\boldsymbol{\eta}η. The vielbein components for observers in the stationary orthonormal frame are
( e t ) t ^ = e Φ ( e r ) r ^ = e Λ ( e θ ) θ ^ = r ( e ϕ ) ϕ ^ = r sin θ e t t ^ = e Φ e r r ^ = e Λ e θ θ ^ = r e ϕ ϕ ^ = r sin θ {:[(e_(t))^( hat(t))=e^(Phi)],[(e_(r))^( hat(r))=e^(Lambda)],[(e_(theta))^( hat(theta))=r],[(e_(phi))^( hat(phi))=r sin theta]:}\begin{aligned} \left(\boldsymbol{e}_{t}\right)^{\hat{t}} & =\mathrm{e}^{\Phi} \\ \left(\boldsymbol{e}_{r}\right)^{\hat{r}} & =\mathrm{e}^{\Lambda} \\ \left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}} & =r \\ \left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}} & =r \sin \theta \end{aligned}(et)t^=eΦ(er)r^=eΛ(eθ)θ^=r(eϕ)ϕ^=rsinθ
In the Schwarzschild case, we can safely call the observer in the this frame stationary, since we can say they are stationary relative to the mass at the origin.
9 9 ^(9){ }^{9}9 See the exercises at the end of Chapter 11 and also Chapter 36 .
10 10 ^(10){ }^{10}10 The simplest way to compute these is to use the useful rules that
G 0 0 = ( R 12 12 + R 23 23 + R 31 31 ) G 0 0 = R 12 12 + R 23 23 + R 31 31 G_(0)^(0)=-(R^(12)_(12)+R^(23)_(23)+R^(31)_(31))G_{0}^{0}=-\left(R^{12}{ }_{12}+R^{23}{ }_{23}+R^{31}{ }_{31}\right)G00=(R1212+R2323+R3131),
G 1 1 = ( R 02 02 + R 03 03 + R 23 23 ) G 1 1 = R 02 02 + R 03 03 + R 23 23 G^(1)_(1)=-(R^(02)_(02)+R^(03)_(03)+R^(23)_(23))G^{1}{ }_{1}=-\left(R^{02}{ }_{02}+R^{03}{ }_{03}+R^{23}{ }_{23}\right)G11=(R0202+R0303+R2323),
G 0 1 = R 02 12 + R 03 13 G 0 1 = R 02 12 + R 03 13 G^(0)_(1)=R^(02)_(12)+R^(03)_(13)G^{0}{ }_{1}=R^{02}{ }_{12}+R^{03}{ }_{13}G01=R0212+R0313,
G 1 2 = R 10 20 + R 13 23 G 1 2 = R 10 20 + R 13 23 G^(1)_(2)=R^(10)_(20)+R^(13)_(23)G^{1}{ }_{2}=R^{10}{ }_{20}+R^{13}{ }_{23}G12=R1020+R1323,
where other components can be found using cyclic permutations. You're invited to prove these rules in Exercise 21.3.
This gives us the left-hand (geometrical) side of the Einstein equation. In the next section, we look at the right-hand (physical) side.
11 11 ^(11){ }^{11}11 That is, in the orthonormal frame with basis vectors e t ^ , e r ^ , e θ ^ , e ϕ ^ e t ^ , e r ^ , e θ ^ , e ϕ ^ e_( hat(t)),e_( hat(r)),e_( hat(theta)),e_( hat(phi))\boldsymbol{e}_{\hat{t}}, \boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}, \boldsymbol{e}_{\hat{\phi}}et^,er^,eθ^,eϕ^ we have
(21.11) T μ ^ ν ^ = ( ρ 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p ) (21.11) T μ ^ ν ^ = ρ 0 0 0 0 p 0 0 0 0 p 0 0 0 0 p {:(21.11)T^( hat(mu) hat(nu))=([rho,0,0,0],[0,p,0,0],[0,0,p,0],[0,0,0,p]):}T^{\hat{\mu} \hat{\nu}}=\left(\begin{array}{cccc} \rho & 0 & 0 & 0 \tag{21.11}\\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{array}\right)(21.11)Tμ^ν^=(ρ0000p0000p0000p)
12 12 ^(12){ }^{12}12 This implies that
(21.13) e 2 Λ = ( 1 2 m ( r ) r ) 1 (21.13) e 2 Λ = 1 2 m ( r ) r 1 {:(21.13)e^(2Lambda)=(1-(2m(r))/(r))^(-1):}\begin{equation*} \mathrm{e}^{2 \Lambda}=\left(1-\frac{2 m(r)}{r}\right)^{-1} \tag{21.13} \end{equation*}(21.13)e2Λ=(12m(r)r)1

21.3 A gravitating object

Now for the right-hand side of the Einstein equation. We make a spherically symmetric distribution of static perfect fluid so that, in the orthonormal frame, the inside of the mass distribution is described by T μ ^ ν ^ = diag ( ρ , p , p , p ) T μ ^ ν ^ = diag ( ρ , p , p , p ) T^( hat(mu) hat(nu))=diag(rho,p,p,p)T^{\hat{\mu} \hat{\nu}}=\operatorname{diag}(\rho, p, p, p)Tμ^ν^=diag(ρ,p,p,p), where ρ ρ rho\rhoρ is the mass density and p p ppp is the pressure. 11 11 ^(11){ }^{11}11 We'll take the total mass of the distribution giving rise to the metric to be M M MMM, and stipulate that the mass distribution stretches out to some maximum radius r = R r = R r=Rr=Rr=R. Outside this radius there is no matter and so the components of T T T\boldsymbol{T}T must vanish locally.
The function Λ ( r ) Λ ( r ) Lambda(r)\Lambda(r)Λ(r) can be determined by considering the 00th component of the Einstein equation. We have, from the previous section, that the component G t ^ t ^ = 8 π T t ^ t ^ G t ^ t ^ = 8 π T t ^ t ^ G_( hat(t) hat(t))=8piT_( hat(t) hat(t))G_{\hat{t} \hat{t}}=8 \pi T_{\hat{t} \hat{t}}Gt^t^=8πTt^t^ can be written as
(21.12) 1 r 2 d d r [ r ( 1 e 2 Λ ) ] = 8 π ρ (21.12) 1 r 2 d d r r 1 e 2 Λ = 8 π ρ {:(21.12)(1)/(r^(2))*((d))/((d)r)*[r(1-e^(-2Lambda))]=8pi rho:}\begin{equation*} \frac{1}{r^{2}} \cdot \frac{\mathrm{~d}}{\mathrm{~d} r} \cdot\left[r\left(1-\mathrm{e}^{-2 \Lambda}\right)\right]=8 \pi \rho \tag{21.12} \end{equation*}(21.12)1r2 d dr[r(1e2Λ)]=8πρ
We will see very shortly that it makes sense to call the quantity in the square brackets 2 m ( r ) 2 m ( r ) 2m(r)2 m(r)2m(r), which is to say 12 12 ^(12){ }^{12}12
(21.14) r ( 1 e 2 Λ ) = 2 m ( r ) (21.14) r 1 e 2 Λ = 2 m ( r ) {:(21.14)r(1-e^(-2Lambda))=2m(r):}\begin{equation*} r\left(1-\mathrm{e}^{-2 \Lambda}\right)=2 m(r) \tag{21.14} \end{equation*}(21.14)r(1e2Λ)=2m(r)
Returning to the Einstein equation 21.12, this definition of m ( r ) m ( r ) m(r)m(r)m(r) gives us
(21.15) 2 r 2 d m ( r ) d r = 8 π ρ (21.15) 2 r 2 d m ( r ) d r = 8 π ρ {:(21.15)(2)/(r^(2))*((d)m(r))/(dr)=8pi rho:}\begin{equation*} \frac{2}{r^{2}} \cdot \frac{\mathrm{~d} m(r)}{\mathrm{d} r}=8 \pi \rho \tag{21.15} \end{equation*}(21.15)2r2 dm(r)dr=8πρ
whose solution is
(21.16) m ( r ) = 0 r d r 4 π r 2 ρ + m ( 0 ) (21.16) m ( r ) = 0 r d r 4 π r 2 ρ + m ( 0 ) {:(21.16)m(r)=int_(0)^(r)dr4pir^(2)rho+m(0):}\begin{equation*} m(r)=\int_{0}^{r} \mathrm{~d} r 4 \pi r^{2} \rho+m(0) \tag{21.16} \end{equation*}(21.16)m(r)=0r dr4πr2ρ+m(0)
Since ρ ρ rho\rhoρ is a density, this expression makes physical sense for a spherically symmetrical object if we interpret m ( r ) m ( r ) m(r)m(r)m(r) to be the mass contained in a sphere of radius r r rrr.
In order to find Φ ( r ) Φ ( r ) Phi(r)\Phi(r)Φ(r), we need only consider the r ^ r ^ r ^ r ^ hat(r) hat(r)\hat{r} \hat{r}r^r^ component of the Einstein equation, which gives us
(21.17) 1 r 2 + 1 r 2 e 2 Λ + 2 r e 2 Λ d Φ d r = G r ^ r ^ = 8 π p (21.17) 1 r 2 + 1 r 2 e 2 Λ + 2 r e 2 Λ d Φ d r = G r ^ r ^ = 8 π p {:(21.17)-(1)/(r^(2))+(1)/(r^(2))e^(-2Lambda)+(2)/(r)e^(-2Lambda)((d)Phi)/((d)r)=G_( hat(r) hat(r))=8pi p:}\begin{equation*} -\frac{1}{r^{2}}+\frac{1}{r^{2}} \mathrm{e}^{-2 \Lambda}+\frac{2}{r} \mathrm{e}^{-2 \Lambda} \frac{\mathrm{~d} \Phi}{\mathrm{~d} r}=G_{\hat{r} \hat{r}}=8 \pi p \tag{21.17} \end{equation*}(21.17)1r2+1r2e2Λ+2re2Λ dΦ dr=Gr^r^=8πp
Substituting for Λ Λ Lambda\LambdaΛ with m ( r ) m ( r ) m(r)m(r)m(r) (from eqn 21.13), this expression becomes 13 13 ^(13){ }^{13}13
(21.19) d Φ d r = m ( r ) + 4 π r 3 p r [ r 2 m ( r ) ] (21.19) d Φ d r = m ( r ) + 4 π r 3 p r [ r 2 m ( r ) ] {:(21.19)(dPhi)/((d)r)=(m(r)+4pir^(3)p)/(r[r-2m(r)]):}\begin{equation*} \frac{\mathrm{d} \Phi}{\mathrm{~d} r}=\frac{m(r)+4 \pi r^{3} p}{r[r-2 m(r)]} \tag{21.19} \end{equation*}(21.19)dΦ dr=m(r)+4πr3pr[r2m(r)]
This looks complicated, but if we concentrate on the field outside of the mass distribution, were p = 0 p = 0 p=0p=0p=0, then we can solve eqn 21.19 straightforwardly. In the region r > R r > R r > Rr>Rr>R, we deduce from eqn 21.16 that the mass function m ( r > R ) m ( r > R ) m(r > R)m(r>R)m(r>R) must be constant. This constant must, of course, be the total mass M M MMM and so we exchange m ( r ) m ( r ) m(r)m(r)m(r) for M M MMM, set p = 0 p = 0 p=0p=0p=0 and write an expression valid for outside the mass distribution of
(21.20) d Φ d r = M r ( r 2 M ) (21.20) d Φ d r = M r ( r 2 M ) {:(21.20)(dPhi)/((d)r)=(M)/(r(r-2M)):}\begin{equation*} \frac{\mathrm{d} \Phi}{\mathrm{~d} r}=\frac{M}{r(r-2 M)} \tag{21.20} \end{equation*}(21.20)dΦ dr=Mr(r2M)
The solution to the latter expression is 14 14 ^(14){ }^{14}14
(21.22) Φ ( r ) = 1 2 ln ( 1 2 M r ) (outside the mass). (21.22) Φ ( r ) = 1 2 ln 1 2 M r  (outside the mass).  {:(21.22)Phi(r)=(1)/(2)ln(1-(2M)/(r))quad" (outside the mass). ":}\begin{equation*} \Phi(r)=\frac{1}{2} \ln \left(1-\frac{2 M}{r}\right) \quad \text { (outside the mass). } \tag{21.22} \end{equation*}(21.22)Φ(r)=12ln(12Mr) (outside the mass). 
The resulting expressions for e 2 Λ e 2 Λ e^(2Lambda)\mathrm{e}^{2 \Lambda}e2Λ and e 2 Φ e 2 Φ e^(2Phi)\mathrm{e}^{2 \Phi}e2Φ give us all we need to justify the form of the Schwarzschild line element outside a gravitating body, which is 15 15 ^(15){ }^{15}15
(21.23) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.23) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.23)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.23} \end{equation*}(21.23)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)

Example 21.2

Substituting the more familiar variables M M MMM and r r rrr, we find
Φ = 1 2 ln ( 1 2 M r ) , Φ = M r ( r 2 M ) , Φ = 2 M ( r M ) r 2 ( r 2 M ) 2 Λ = 1 2 ln ( r r 2 M ) , Λ = M r ( r 2 M ) Φ = 1 2 ln 1 2 M r , Φ = M r ( r 2 M ) , Φ = 2 M ( r M ) r 2 ( r 2 M ) 2 Λ = 1 2 ln r r 2 M , Λ = M r ( r 2 M ) {:[Phi=(1)/(2)ln(1-(2M)/(r))",",Phi^(')=(M)/(r(r-2M))",",Phi^('')=(-2M(r-M))/(r^(2)(r-2M)^(2))],[Lambda=(1)/(2)ln((r)/(r-2M))",",Lambda^(')=(-M)/(r(r-2M))]:}\begin{array}{lll} \Phi=\frac{1}{2} \ln \left(1-\frac{2 M}{r}\right), & \Phi^{\prime}=\frac{M}{r(r-2 M)}, & \Phi^{\prime \prime}=\frac{-2 M(r-M)}{r^{2}(r-2 M)^{2}} \\ \Lambda=\frac{1}{2} \ln \left(\frac{r}{r-2 M}\right), & \Lambda^{\prime}=\frac{-M}{r(r-2 M)} \end{array}Φ=12ln(12Mr),Φ=Mr(r2M),Φ=2M(rM)r2(r2M)2Λ=12ln(rr2M),Λ=Mr(r2M)
and also
(21.25) E = 2 M 2 3 , E ¯ = M r 3 , F = 2 M r 3 , F ¯ = r 3 r 3 (21.25) E = 2 M 2 3 , E ¯ = M r 3 , F = 2 M r 3 , F ¯ = r 3 r 3 {:(21.25){:[E=(2M)/(2^(3))",", bar(E)=(-M)/(r^(3))","],[F=(2M)/(r^(3))",", bar(F)=(r^(3))/(r^(3))]:}:}\begin{array}{ll} E=\frac{2 M}{2^{3}}, & \bar{E}=\frac{-M}{r^{3}}, \tag{21.25}\\ F=\frac{2 M}{r^{3}}, & \bar{F}=\frac{r^{3}}{r^{3}} \end{array}(21.25)E=2M23,E¯=Mr3,F=2Mr3,F¯=r3r3
In terms of the familiar variables M M MMM and r r rrr, the useful components of the Riemann curvature tensor are 16 16 ^(16){ }^{16}16
(21.26) R t ^ r ^ r ^ r ^ = 2 M r 3 , R t ^ θ ^ t ^ θ ^ = M r 3 , R t ^ ϕ ^ t ^ ϕ ^ = M r 3 R θ ^ ϕ ^ θ ^ ϕ ^ = 2 M r 3 , R r ^ ϕ ^ r ^ ϕ ^ ϕ ^ = M r 3 , R r ^ θ ^ r ^ θ ^ = M r 3 (21.26) R t ^ r ^ r ^ r ^ = 2 M r 3 , R t ^ θ ^ t ^ θ ^ = M r 3 , R t ^ ϕ ^ t ^ ϕ ^ = M r 3 R θ ^ ϕ ^ θ ^ ϕ ^ = 2 M r 3 , R r ^ ϕ ^ r ^ ϕ ^ ϕ ^ = M r 3 , R r ^ θ ^ r ^ θ ^ = M r 3 {:[(21.26)R_( hat(t) hat(r) hat(r) hat(r))=-(2M)/(r^(3))","quadR_( hat(t) hat(theta) hat(t) hat(theta))=(M)/(r^(3))","quadR_( hat(t) hat(phi) hat(t) hat(phi))=(M)/(r^(3))],[R_( hat(theta) hat(phi) hat(theta) hat(phi))=(2M)/(r^(3))","quadR_( hat(r) hat(phi) hat(r) hat(phi) hat(phi))=-(M)/(r^(3))","quadR_( hat(r) hat(theta) hat(r) hat(theta))=-(M)/(r^(3))]:}\begin{gather*} R_{\hat{t} \hat{r} \hat{r} \hat{r}}=-\frac{2 M}{r^{3}}, \quad R_{\hat{t} \hat{\theta} \hat{t} \hat{\theta}}=\frac{M}{r^{3}}, \quad R_{\hat{t} \hat{\phi} \hat{t} \hat{\phi}}=\frac{M}{r^{3}} \tag{21.26}\\ R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}}=\frac{2 M}{r^{3}}, \quad R_{\hat{r} \hat{\phi} \hat{r} \hat{\phi} \hat{\phi}}=-\frac{M}{r^{3}}, \quad R_{\hat{r} \hat{\theta} \hat{r} \hat{\theta}}=-\frac{M}{r^{3}} \end{gather*}(21.26)Rt^r^r^r^=2Mr3,Rt^θ^t^θ^=Mr3,Rt^ϕ^t^ϕ^=Mr3Rθ^ϕ^θ^ϕ^=2Mr3,Rr^ϕ^r^ϕ^ϕ^=Mr3,Rr^θ^r^θ^=Mr3
We also observe that components of G G G\boldsymbol{G}G vanish as they must, since we're in a region with ρ = p = 0 ρ = p = 0 rho=p=0\rho=p=0ρ=p=0 and so 0 = 8 π T = G 0 = 8 π T = G 0=8pi T=G0=8 \pi \boldsymbol{T}=\boldsymbol{G}0=8πT=G.

Example 21.3

The arguments in this section can also be used to derive the Tolman-Oppenheimer-Volkov (TOV) equations that describe the structure of a static, spherical perfect fluid, and which therefore provides us with a relativistic model of a star. Starting with the stress-energy tensor for a perfect fluid, we can use T = 0 T = 0 grad*T=0\boldsymbol{\nabla} \cdot \boldsymbol{T}=0T=0 to show 17 17 ^(17){ }^{17}17 that in the Schwarzschild geometry
(21.27) p r + ( p + ρ ) Φ r = 0 (21.27) p r + ( p + ρ ) Φ r = 0 {:(21.27)(del p)/(del r)+(p+rho)(del Phi)/(del r)=0:}\begin{equation*} \frac{\partial p}{\partial r}+(p+\rho) \frac{\partial \Phi}{\partial r}=0 \tag{21.27} \end{equation*}(21.27)pr+(p+ρ)Φr=0
This equation can be combined with eqn 21.19 to give the first TOV equation
(21.28) p r = [ p ( r ) + ρ ( r ) ] [ m ( r ) + 4 π r 3 p ( r ) ] r 2 [ 1 2 m ( r ) r ] (21.28) p r = [ p ( r ) + ρ ( r ) ] m ( r ) + 4 π r 3 p ( r ) r 2 1 2 m ( r ) r {:(21.28)(del p)/(del r)=([p(r)+rho(r)][m(r)+4pir^(3)p(r)])/(r^(2)[1-(2m(r))/(r)]):}\begin{equation*} \frac{\partial p}{\partial r}=\frac{[p(r)+\rho(r)]\left[m(r)+4 \pi r^{3} p(r)\right]}{r^{2}\left[1-\frac{2 m(r)}{r}\right]} \tag{21.28} \end{equation*}(21.28)pr=[p(r)+ρ(r)][m(r)+4πr3p(r)]r2[12m(r)r]
The second TOV equation gives us a quantity to be interpreted as the mass
(21.29) m ( r ) = 4 π 0 r d r r 2 ρ ( r ) (21.29) m ( r ) = 4 π 0 r d r r 2 ρ ( r ) {:(21.29)m(r)=4piint_(0)^(r)drr^(2)rho(r):}\begin{equation*} m(r)=4 \pi \int_{0}^{r} \mathrm{~d} r r^{2} \rho(r) \tag{21.29} \end{equation*}(21.29)m(r)=4π0r drr2ρ(r)
where we have fixed the integration constant m ( 0 ) = 0 m ( 0 ) = 0 m(0)=0m(0)=0m(0)=0. The solution to these equations requires a choice of equation of state (that is, a link between p p ppp and ρ ρ rho\rhoρ ). They are generally integrated numerically from the origin outwards until p ( r = R ) = 0 p ( r = R ) = 0 p(r=R)=0p(r=R)=0p(r=R)=0, which we take to give the surface of the star, with m ( R ) m ( R ) m(R)m(R)m(R) giving us the stellar mass.
14 14 ^(14){ }^{14}14 This gives us
(21.21) e 2 Φ ( r ) = ( 1 2 M r ) (21.21) e 2 Φ ( r ) = 1 2 M r {:(21.21)e^(2Phi(r))=(1-(2M)/(r)):}\begin{equation*} \mathrm{e}^{2 \Phi(r)}=\left(1-\frac{2 M}{r}\right) \tag{21.21} \end{equation*}(21.21)e2Φ(r)=(12Mr)
The connection coefficients for the Schwarzschild metric are
Γ t r t = Φ , Γ r t t = Φ e 2 Φ e 2 Λ , Γ r r r r = Λ , Γ r θ θ = r e 2 Λ , Γ r ϕ ϕ = r e 2 Λ sin 2 θ , Γ θ θ Γ θ r θ = 1 r , Γ ϕ ϕ ϕ = sin θ cos θ , Γ ϕ ϕ ϕ = 1 r , cos θ sin θ , Γ t r t = Φ , Γ r t t = Φ e 2 Φ e 2 Λ , Γ r r r r = Λ , Γ r θ θ = r e 2 Λ , Γ r ϕ ϕ = r e 2 Λ sin 2 θ , Γ θ θ Γ θ r θ = 1 r , Γ ϕ ϕ ϕ = sin θ cos θ , Γ ϕ ϕ ϕ = 1 r , cos θ sin θ , {:[Gamma^(t)_(rt)=Phi^(')","],[Gamma^(r)_(tt)=Phi^(')e^(2Phi)e^(-2Lambda)","],[Gamma^(r)r_(rr)=Lambda^(')","],[Gamma^(r)^(theta theta)=-re^(-2Lambda)","],[Gamma^(r)^(phi phi)=-re^(-2Lambda)sin^(2)theta","],[Gamma^(theta)theta],[Gamma^(theta)^(r theta)=(1)/(r)","],[Gamma_(phi phi)^(phi)=-sin theta cos theta","],[Gamma^(phi)^(^(phi)phi)=(1)/(r)","],[cos theta],[sin theta","]:}\begin{aligned} & \Gamma^{t}{ }_{r t}=\Phi^{\prime}, \\ & \Gamma^{r}{ }_{t t}=\Phi^{\prime} \mathrm{e}^{2 \Phi} \mathrm{e}^{-2 \Lambda}, \\ & \Gamma^{r} r_{r r}=\Lambda^{\prime}, \\ & \Gamma^{r}{ }^{\theta \theta}=-r \mathrm{e}^{-2 \Lambda}, \\ & \Gamma^{r}{ }^{\phi \phi}=-r \mathrm{e}^{-2 \Lambda} \sin ^{2} \theta, \\ & \Gamma^{\theta} \theta \\ & \Gamma^{\theta}{ }^{r \theta}=\frac{1}{r}, \\ & \Gamma_{\phi \phi}^{\phi}=-\sin \theta \cos \theta, \\ & \Gamma^{\phi}{ }^{{ }^{\phi} \phi}=\frac{1}{r}, \\ & \cos \theta \\ & \sin \theta, \end{aligned}Γtrt=Φ,Γrtt=Φe2Φe2Λ,Γrrrr=Λ,Γrθθ=re2Λ,Γrϕϕ=re2Λsin2θ,ΓθθΓθrθ=1r,Γϕϕϕ=sinθcosθ,Γϕϕϕ=1r,cosθsinθ,
where a dash denotes a derivative with respect to the r r rrr coordinate.
(These are computed in Exercise 9.7).
15 15 ^(15){ }^{15}15 In a letter to Schwarzschild in 1916, Einstein wrote "I had not expected that one could formulate the exact solution of the problem so simply. The analytcal treatment of the problem seems to me to be excellent."
16 16 ^(16){ }^{16}16 Achieved by lowering indices using the Minkowski tensor (i.e. lowering a t ^ t ^ hat(t)\hat{t}t^ gives a minus sign, lowering everything else has no effect).
Richard C. Tolman (1881-1948) J. Robert Oppenheimer (1904-1967). George Volkoff (1914-2000).
Oppenheimer and Volkoff used Tolman's work as a basis that led to their prediction of the existence of neutron stars.
17 17 ^(17){ }^{17}17 See Exercise 39.5.
18 18 ^(18){ }^{18}18 As examined in the exercises, the Newtonian prediction for hydrostatic equilibrium is
(21.31) 4 π r 2 d p = m d m r 2 (21.31) 4 π r 2 d p = m d m r 2 {:(21.31)4pir^(2)dp=-(m(d)m)/(r^(2)):}\begin{equation*} 4 \pi r^{2} \mathrm{~d} p=-\frac{m \mathrm{~d} m}{r^{2}} \tag{21.31} \end{equation*}(21.31)4πr2 dp=m dmr2
It will be useful in later chapters to have in mind a simple picture of stellar evolution. Stars are formed from gas clouds that collapse under gravity to eventually achieve equilibrium where gravitational collapse is balanced against outward radiation pressure resulting from nuclear fusion of hydrogen in the star's core. The hydrogen in the core eventually becomes exhausted, causing many stars (i.e. those of around one solar mass M M M_(o.)M_{\odot}M ), to fuse helium inside their core and hydrogen outside, leading to expansion into a red giant leading the nuclear fuel is a red giant Once the mare fuel is exhausted, the star collapses under its own gravity typically forming a white dwarf, with outer layers of mass thrown off as planetary nebula. The white dwarf achieves equilibrium with gravitation collapse now balanced by the outward pressure caused by the electronic matter making up the dwarf being degenerate. The latter implies the density of electronic matter is such that the quantum energy levels are completely occupied by electrons, causing them to stack up in energy owing to the Pauli principle. This leads to a large outward pressure reflecting the difficulty in rearranging particles between energy levels while preventing multiple occupancy. For stars with masses 1.5 M 1.5 M ≳1.5M_(o.)\gtrsim 1.5 M_{\odot}1.5M, the white dwarf cannot achieve this equilibrium owing to the size of the inward gravitational force, and a neutron star or black hole can be formed. The former involves electrons and protons fusing to form neutrons, with equilibrium now achieved from the outward pressure of the degenerate neutron matter. Pulsars were discovered by Jocelyn Bell Burnell (1943-) in 1967, and were later identified as rapidly rotating neutron stars Black holes are discussed in Chapter 25.
19 19 ^(19){ }^{19}19 We will often invoke the observer at infinity, whose proper time τ = t τ = t tau=t\tau=tτ=t.
To interpret the first equation it can be rewritten as
(21.30) 4 π r 2 d p ( r ) = m ( r ) d m r 2 [ 1 + p ( r ) ρ ( r ) ] [ 1 + 4 π r 3 p ( r ) m ( r ) ] [ 1 2 m ( r ) r ] 1 (21.30) 4 π r 2 d p ( r ) = m ( r ) d m r 2 1 + p ( r ) ρ ( r ) 1 + 4 π r 3 p ( r ) m ( r ) 1 2 m ( r ) r 1 {:(21.30)4pir^(2)dp(r)=-(m(r)dm)/(r^(2))[1+(p(r))/(rho(r))][1+(4pir^(3)p(r))/(m(r))][1-(2m(r))/(r)]^(-1):}\begin{equation*} 4 \pi r^{2} \mathrm{~d} p(r)=-\frac{m(r) \mathrm{d} m}{r^{2}}\left[1+\frac{p(r)}{\rho(r)}\right]\left[1+\frac{4 \pi r^{3} p(r)}{m(r)}\right]\left[1-\frac{2 m(r)}{r}\right]^{-1} \tag{21.30} \end{equation*}(21.30)4πr2 dp(r)=m(r)dmr2[1+p(r)ρ(r)][1+4πr3p(r)m(r)][12m(r)r]1
The first term on the right is the Newtonian prediction 18 18 ^(18){ }^{18}18 with the remaining terms giving the relativistic corrections. In low-mass stars, the largest contribution to ρ ρ rho\rhoρ is from baryons (i.e. nuclear matter), which don't contribute significantly to the pressure (which turns out to be provided by electrons). This means p / ρ 0 p / ρ 0 p//rho~~0p / \rho \approx 0p/ρ0 and p / m 0 p / m 0 p//m~~0p / m \approx 0p/m0, so that the relativistic corrections are not large. In larger stars, the pressure and energy density increase the right-hand side of the equation, steepening the pressure energy density increase the right-hand side of the equation, steepening the pressure the prediction of Newtonian physics

21.4 The meaning of the coordinates

In general relativity, coordinates have no intrinsic metric significance. However, we can relate the coordinates to the quantities of interest in describing stars and orbits in this specific case. This is our next task.

Example 21.4

Let's consider a circle in the equatorial plane ( θ = π / 2 ) ( θ = π / 2 ) (theta=pi//2)(\theta=\pi / 2)(θ=π/2) at an instant in time ( d t = 0 d t = 0 dt=0\mathrm{d} t=0dt=0 ) We then have
(21.32) d s 2 = r 2 d ϕ 2 . (21.32) d s 2 = r 2 d ϕ 2 . {:(21.32)ds^(2)=r^(2)dphi^(2).:}\begin{equation*} \mathrm{d} s^{2}=r^{2} \mathrm{~d} \phi^{2} . \tag{21.32} \end{equation*}(21.32)ds2=r2 dϕ2.
Taking a square root and integrating, we find the circumference of the circle is
(21.33) C = ϕ = 0 2 π r d ϕ = 2 π r (21.33) C = ϕ = 0 2 π r d ϕ = 2 π r {:(21.33)C=int_(phi=0)^(2pi)rdphi=2pi r:}\begin{equation*} C=\int_{\phi=0}^{2 \pi} r \mathrm{~d} \phi=2 \pi r \tag{21.33} \end{equation*}(21.33)C=ϕ=02πr dϕ=2πr
We can therefore call r r rrr a radius, in that it supplies the correct scaling to work out the circumference of circles. We must be careful, since if we attempt to compute the distance between two events along a radial line, we obtain
(21.34) Δ s = r A r B ( 1 2 M r ) 1 2 d r (21.34) Δ s = r A r B 1 2 M r 1 2 d r {:(21.34)Delta s=int_(r_(A))^(r_(B))(1-(2M)/(r))^(-(1)/(2))dr:}\begin{equation*} \Delta s=\int_{r_{\mathrm{A}}}^{r_{\mathrm{B}}}\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \mathrm{~d} r \tag{21.34} \end{equation*}(21.34)Δs=rArB(12Mr)12 dr
which will be greater than r B r A r B r A r_(B)-r_(A)r_{\mathrm{B}}-r_{\mathrm{A}}rBrA.
We conclude that r r rrr does not give the distance from the origin in the Schwarzschild geometry. Since it does give the radius of circles it is sometimes known as the circumferential radial coordinate.
Next, we turn to the time. We find that the proper time between two events is
(21.35) Δ τ = ( 1 2 M r ) Δ t (21.35) Δ τ = 1 2 M r Δ t {:(21.35)Delta tau=(1-(2M)/(r))Delta t:}\begin{equation*} \Delta \tau=\left(1-\frac{2 M}{r}\right) \Delta t \tag{21.35} \end{equation*}(21.35)Δτ=(12Mr)Δt
For r r r rarr oor \rightarrow \inftyr we have Δ τ = Δ t Δ τ = Δ t Delta tau=Delta t\Delta \tau=\Delta tΔτ=Δt. We conclude that the t t ttt coordinate is the time between events as measured by a clock at infinity. 19 19 ^(19){ }^{19}19
We now have a suitable metric field we can use it to examine the motion that is possible in the curved spacetime that it describes. In the next chapter, we look at general properties of motion in the Schwarzschild geometry before, in Chapter 23 , turning to the question of orbits.

Chapter summary

  • The Schwarzschild geometry refers to static, spherically symmetric spacetime.
  • The metric field in the Schwarzschild geometry is given by the Schwarzschild line element
(21.36) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (21.36) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(21.36)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.36} \end{equation*}(21.36)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
  • The Schwarzschild metric is static, asymptotically flat and is badly behaved at r = 0 r = 0 r=0r=0r=0 and r = 2 M r = 2 M r=2Mr=2 Mr=2M.
  • Birkhoff's theorem says that any spherically symmetric solution to Einstein's equation is identical to the Schwarzschild solution.

Exercises

(21.1) Use the vielbein components to express the components of the energy-momentum tensor T T T\boldsymbol{T}T for a perfect fluid in a ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ) coordinate system.
(21.2) (a) Verify eqns 21.10 using the method suggested in the text.
(b) Using eqns 21.25 , write the components of G G GGG in the orthonormal frame using the familiar polar coordinates and show that each component vanishes.
(21.3) Using the symmetries of R R R\boldsymbol{R}R, prove the useful rule in Sidenote 10.
(21.4) (a) Confirm that eqn 21.19 follows from eqn 21.17. (b) Show that eqn 21.19 implies eqn 21.18 in the Newtonian limit.
(21.5) (a) By computing the relevant determinant g g ggg, compute the area of a spherical surface of fixed r r rrr and t t ttt in the Schwarzschild geometry.
(b) Compute a circumference at fixed r , t r , t r,tr, tr,t and θ θ theta\thetaθ.
(21.6) (a) Show that in Newtonian gravity, hydrostatic equilibrium requires that
(21.37) 4 π r 2 d p = G M d m r 2 (21.37) 4 π r 2 d p = G M d m r 2 {:(21.37)4pir^(2)dp=-(GM(d)m)/(r^(2)):}\begin{equation*} 4 \pi r^{2} \mathrm{~d} p=-\frac{G M \mathrm{~d} m}{r^{2}} \tag{21.37} \end{equation*}(21.37)4πr2 dp=GM dmr2
where d m = 4 π r 2 ρ d r d m = 4 π r 2 ρ d r dm=4pir^(2)rhodr\mathrm{d} m=4 \pi r^{2} \rho \mathrm{~d} rdm=4πr2ρ dr.
Hint: Generalize the usual derivation of Pascal's
law of hydrostatics in a uniform gravitational field for the case of a non-uniform field.
(21.7) Justification of Birkhoff's theorem. The most general spherically symmetric line element has the form
d s 2 = A ( r , t ) d t 2 + B ( r , t ) d r d t + 2 C ( r , t ) d r 2 (21.38) + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = A ( r , t ) d t 2 + B ( r , t ) d r d t + 2 C ( r , t ) d r 2 (21.38) + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-A(r","t)dt^(2)+B(r","t)drdt+2C(r","t)dr^(2)],[(21.38)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{align*} \mathrm{d} s^{2}= & -A(r, t) \mathrm{d} t^{2}+B(r, t) \mathrm{d} r \mathrm{~d} t+2 C(r, t) \mathrm{d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{21.38} \end{align*}ds2=A(r,t)dt2+B(r,t)dr dt+2C(r,t)dr2(21.38)+r2( dθ2+sin2θ dϕ2)
(a) Show that a transformation t t + h ( r , t ) t t + h ( r , t ) t rarr t+h(r,t)t \rightarrow t+h(r, t)tt+h(r,t), where h ( r , t ) h ( r , t ) h(r,t)h(r, t)h(r,t) is some function, can be used to eliminate the cross term proportional to d r d t d r d t drdt\mathrm{d} r \mathrm{~d} tdr dt, such that we have
d s 2 = e Φ ( r , t ) d t 2 + e Λ ( r , t ) d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = e Φ ( r , t ) d t 2 + e Λ ( r , t ) d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-e^(Phi(r,t))dt^(2)+e^(Lambda(r,t))dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))\mathrm{d} s^{2}=-\mathrm{e}^{\Phi(r, t)} \mathrm{d} t^{2}+\mathrm{e}^{\Lambda(r, t)} \mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)ds2=eΦ(r,t)dt2+eΛ(r,t)dr2+r2( dθ2+sin2θ dϕ2). (21.39)
In Exercise 36.8, we compute the curvature properties of this spacetime. It follows from these that the components of the Einstein tensor are
$G_{\hat{t} \hat{t}}=r^{-2}\left(1-\mathrm{e}^{-2 \Lambda}\right)+2(\Lambda, r / r) \mathrm{e}^{-2 \Lambda}$,
$G_{\hat{r} \hat{t}}=2\left(\Lambda_{t} / r\right) \mathrm{e}^{-(\Lambda+\Phi)}$,
$G_{\hat{r} \hat{r}}=2\left(\Phi_{, r} / r\right) \mathrm{e}^{-2 \Lambda}+r^{-2}\left(\mathrm{e}^{-2 \Lambda}-1\right)$,
$G_{\hat{\theta} \hat{\theta}}=G_{\hat{\phi} \hat{\phi}},=$
        $\left(\Phi_{, r r}+\Phi_{, r}^{2}-\Phi_{, r} \Lambda_{, r}+\Phi_{, r} / r-\Lambda_{, r} / r\right) \mathrm{e}^{-2 \Lambda}$
        $-\left(\Lambda_{, t t}+\Lambda_{, t}^{2}-\Phi_{, t} \Lambda_{, t}\right) \mathrm{e}^{-2 \Phi}$.
(b) Show that if we are in a vacuum, then $\Lambda(r, t)$ is independent of time.
(c) Use the remaining components of the Einstein equation to show that
$$
\begin{equation*}
\Lambda=-\frac{1}{2} \ln \left|1-\frac{2 M}{r}\right| \tag{21.41}
\end{equation*}
$$
(d) Show further that
$$
\begin{equation*}
\Phi(r, t)=-\Lambda(r)+f(t) \tag{21.42}
\end{equation*}
$$
where $f(t)$ is a function of time.
(e) Use this to prove Birkhoff's theorem, which says that the Schwarzschild geometry is the most general, asymptotically flat, spherically symmetric solution of the Einstein equation.
(21.8) Consider filling a universe described by the line element in eqn 21.5 with energy density $\zeta$, such that we have $T_{\hat{t} \hat{t}}=8 \pi \zeta$ and $T_{\hat{i} \hat{i}}=-8 \pi \zeta$.
(a) Use the results from this chapter to show that this energy density is consistent with $\Phi=-\Lambda$.
(b) Show that the line element
$$
\begin{align*}
\mathrm{d} s^{2}= & -\left(1-H^{2} r^{2}\right) \mathrm{d} t^{2}+\left(1-H^{2} r^{2}\right)^{-1} \mathrm{~d} r^{2} \\
& +r^{2} \mathrm{~d} \Omega^{2} \tag{21.43}
\end{align*}
$$
is a solution to the Einstein equation. Here $\mathrm{d} \Omega^{2}=$ $\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}$ and $H$ is a constant you should determine.
This solution is actually the de Sitter model (Universe 1 from Chapter 15) again, expressed in a different set of coordinates. The link is made in Exercise 18.8.

\section*{Motion in the Schwarzschild geometry}

We are merely the stars' tennis-balls, struck and bandied Which way to please them
John Webster (1580-1625) The Duchess of Malfi

Thanne longen folk to goon on pilgrimages
Geoffrey Chaucer (c.1343-1400)
Prologue to the Cantebury Tales

We found in the last chapter that the Schwarzschild line element is given by
$$
\begin{equation*}
\mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{22.2}
\end{equation*}
$$

This equation represents the metric field outside a spherically symmetric object, such as a star, of mass $M$. In this chapter and the following ones, we shall examine the free-falling motion of particles in this metric field, where the particles follow the geodesics of this geometry. We will find that general relativity allows a richer range of possible motions than are found in the Newtonian problem, including, in addition to the interesting trajectories of massive particles, effects on the motion of photons: the particles of light.
At first glance the problem of relativistic motion looks like a formidable one involving deriving solutions to the geodesic equation, ${ }^{1}$ with terms provided by the connection coefficients found from eqn 22.2 . However, just as problems involving Newtonian orbits are greatly simplified by knowing about conserved quantities, we shall see how an analysis of conserved quantities, along with simple facts about the velocity $\boldsymbol{u}$, allow us to solve a variety of key problems. In particular, we make a lot of use of the identity ${ }^{2} \boldsymbol{u} \cdot \boldsymbol{u}=-1$ for massive particles and $\boldsymbol{u} \cdot \boldsymbol{u}=0$ for massless ones. In addition, just as energy and angular momentum are conserved in the Newtonian potential, we will see that something very much like these is also conserved in the relativistic case, as we might naively expect. We turn first, therefore, to the method for extracting conserved quantities.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-087.jpg?height=209&width=389&top_left_y=469&top_left_x=1206)
22.1 Constants of the motion 238 22.2 Gravitational redshift 239 22.3 Motion in Schwarzschild spacetime 240 22.4 Example: the radial plunge

Chapter summary 244
Exercises 244

In this chapter, we'll need to use the components of the Schwarzschild metric
$$
\begin{align*}
g_{t t} & =-\left(1-\frac{2 M}{r}\right) \\
g_{r r} & =\left(1-\frac{2 M}{r}\right)^{-1} \\
g_{\theta \theta} & =r^{2} \\
g_{\phi \phi} & =r^{2} \sin ^{2} \theta \tag{22.1}
\end{align*}
$$
${ }^{1}$ This is the differential equation
$$
\frac{\mathrm{d}^{2} x^{\mu}}{\mathrm{d} \tau^{2}}+\Gamma_{\alpha \beta}^{\mu} \frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \tau} \cdot \frac{\mathrm{~d} x^{\beta}}{\mathrm{d} \tau}=0
$$
with the connection coefficients determined from the metric.
${ }^{2}$ Reminder: A massive particle's world line is parametrized by proper time so that, in a coordinate system with $x^{\mu}=$ ( $t, r, \theta, \phi$ ), the velocity $\boldsymbol{u}$ has components $u^{\mu}=\left(u^{t}, u^{r}, u^{\theta}, u^{\phi}\right)$, i.e.
$$
u^{\mu}=\left(\frac{\mathrm{d} t}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \tau}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \theta}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}\right)
$$
$\curvearrowright$ Killing vectors are discussed in detail in Chapter 33 in the context of geometry.

\subsection*{22.1 Constants of the motion}

One of the most important factors in analysing motion in mechanics is the identification of constants of the motion. In geometrical problems, exemplified by the physics of gravitation, conserved quantities can be identified by finding a set of fields known as the Killing vector fields $\boldsymbol{\xi}$ of the metric. These are very simply identified by inspection, by noting which coordinates do not feature in the components of the metric. For now we will give (but not prove) a simple recipe for identifying conserved quantities.
To find a conserved quantity from a metric field:
Step 1: Look at the metric components and identify those coordinates on which none of the metric components depend.
Step 2: If the variable $x^{\alpha}$ does not feature, we have a Killing vector $\boldsymbol{\xi}=\boldsymbol{e}_{\alpha}$.
Step 3: Particles travelling along a geodesic have a velocity $\boldsymbol{u}=u^{\mu} \boldsymbol{e}_{\mu}$, which is the tangent to a geodesic. The quantity
$$
\begin{equation*}
\boldsymbol{\xi} \cdot \boldsymbol{u}=g_{\mu \nu} \xi^{\mu} u^{\nu} \tag{22.3}
\end{equation*}
$$
is conserved along the geodesic.
To identify the Killing vectors for the Schwarzschild geometry, we spot (step 1) that the metric components are all independent of the coordinates $t$ and $\phi$. Ordering the coordinates $(t, r, \theta, \phi)$ we identify two corresponding Killing vectors (step 2). First, from the independence of $t$ we extract a Killing vector $\boldsymbol{\xi}$ via
$$
\begin{equation*}
\boldsymbol{\xi}=\boldsymbol{e}_{t}, \quad \xi^{\mu}=(1,0,0,0), \quad(\text { time independence }) \tag{22.4}
\end{equation*}
$$
and from the independence of $\phi$ we identify a Killing vector $\boldsymbol{\eta}$ as we have
$$
\begin{equation*}
\boldsymbol{\eta}=\boldsymbol{e}_{\phi}, \quad \eta^{\mu}=(0,0,0,1), \quad \text { (rotational symmetry) } \tag{22.5}
\end{equation*}
$$

We now use the fact that $\boldsymbol{\xi} \cdot \boldsymbol{u}$ is a constant along a geodesic (step 3) to identify the conserved quantities. The first is identified $\mathrm{as}^{3}$
$$
\begin{equation*}
\boldsymbol{\xi} \cdot \boldsymbol{u}=u^{\mu} \boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{t}=u^{\mu} g_{\mu t}=u_{t} \tag{22.6}
\end{equation*}
$$
or equivalently
$$
\begin{equation*}
\boldsymbol{\xi} \cdot \boldsymbol{u}=g_{t t} u^{t}=-\left(1-\frac{2 M}{r}\right) u^{t}=-\left(1-\frac{2 M}{r}\right) \frac{\mathrm{d} t}{\mathrm{~d} \tau} \tag{22.7}
\end{equation*}
$$
which is conventionally named $-\tilde{E}$, which is to say we have a conserved quantity
$$
\begin{equation*}
\tilde{E}=-\boldsymbol{\xi} \cdot \boldsymbol{u}=\left(1-\frac{2 M}{r}\right) u^{t} \tag{22.8}
\end{equation*}
$$

This quantity $\tilde{E}$ tells us about the energy per unit rest mass. ${ }^{4}$ In general, it can be interpreted, for timelike geodesics, as total energy per unit rest mass, measured by a static observer at infinity.

The other conserved quantity, called $\tilde{L}$ is also straightforward to pick out as $\boldsymbol{\eta} \cdot \boldsymbol{u}=u_{\phi}$, or, from $u_{\phi}=g_{\phi \phi} u^{\phi}$
$$
\begin{equation*}
\tilde{L}=\boldsymbol{\eta} \cdot \boldsymbol{u}=r^{2} \sin ^{2} \theta u^{\phi} . \tag{22.9}
\end{equation*}
$$

We interpret this as angular momentum per unit rest mass. Note that for equatorial motion (i.e. with $\theta=\pi / 2$ ) this becomes $\tilde{L}=r^{2} \dot{\phi}$, which looks just like the expression for angular momentum per unit mass in cylindrical coordinates.
So far we have considered only the geometry, and not what sort of particle is in free fall. This information is supplied via the square of the velocity vector $\boldsymbol{u} \cdot \boldsymbol{u}$. Massive particles move along timelike geodesics and so we fix the magnitude of the velocity vector (which is tangent to the geodesic) by saying
$$
\begin{equation*}
\boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \cdot \frac{\mathrm{~d} x^{\nu}}{\mathrm{d} \tau}=-1, \quad \text { (massive particles) } \tag{22.10}
\end{equation*}
$$
where the proper time $\tau$ has been employed as an affine parameter marking off the trajectory of the particle. Photons travel along null geodesics, whose velocity tangent vector has the property
$$
\begin{equation*}
\boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} \frac{\mathrm{d} x^{\mu} \lambda}{\mathrm{d} \lambda} \frac{\mathrm{~d} x^{\nu}}{\mathrm{d} \lambda}=0, \quad \text { (massless photons) } \tag{22.11}
\end{equation*}
$$
where $\lambda$ is an affine parameter for the photon world line.

\subsection*{22.2 Gravitational redshift}

With these preliminaries, we can derive some results. The first is a result for photons: the gravitational redshift caused by the Schwarzschild geometry. ${ }^{5}$ To analyse this, we consider the motion of a photon with null momentum $\boldsymbol{p}$ propagating along a radial line. By symmetry, this is a geodesic. The photon has the property ${ }^{6}$ that $\boldsymbol{p} \cdot \boldsymbol{\xi}$ is conserved along the geodesic, which implies that, in this geometry, $g_{t t} p^{t}=(1-2 M / r) p^{t}(r)$ is conserved as $r$ is varied (rather than the flat-space energy $p^{t}$, as we might naively have expected). Remember, however, that an observer at $r$ does not measure $p^{t}$; they measure the local value $p^{\hat{t}}=\hbar \omega$.

We can find out how the photon's measured frequency $\omega$ changes as the photon moves radially outwards from a massive star. We consider two static observers, one at radius $r=R$ and one at $r=\infty$ (Fig. 22.1). Being observers, their world lines have tangents that are timelike velocity vectors. The energy of a photon with momentum $\boldsymbol{p}$, measured by an observer ${ }^{7}$ with velocity $\boldsymbol{u}_{\text {obs }}$ is
$$
\begin{equation*}
\hbar \omega=-\boldsymbol{p} \cdot \boldsymbol{u}_{\mathrm{obs}} \tag{22.12}
\end{equation*}
$$

If the observer is at rest with respect to the star, then $u_{\text {obs }}^{i}=0$ and we can determine the timelike component of $\boldsymbol{u}_{\text {obs }}$ using the velocity identity
$$
\begin{equation*}
g_{t t}\left(u_{\mathrm{obs}}^{t}\right)^{2}=-1 \tag{22.13}
\end{equation*}
$$
$t$
A
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-089.jpg?height=249&width=283&top_left_y=936&top_left_x=1233)

Fig. 22.1 A photon world line meets an observer at $R$ and one at infinity.
${ }^{5}$ Here we revisit our second cause of frequency shifts discussed earlier in Chapter 6. (The others are the Doppler effect and the cosmological redshift.)
${ }^{6}$ This follows since we shall assume $\boldsymbol{p}$ to be parallel to $\boldsymbol{u}$ for light and $\boldsymbol{p}$ is therefore tangent to the geodesics. (For a more careful discussion of photon momentum, see Section 24.7.)
${ }^{7}$ An alternative rule would be to use the vielbein components to say that the measured energy $p^{t}$ is related to $p^{t}$ via
$$
p^{\hat{t}}(r)=\left(\boldsymbol{e}_{\mu}\right)^{\hat{t}}(r) p^{\mu}(r)
$$

This alternative approach is examined in the exercises. Here, instead, since we are trying to relate measurements made by local observers in two places, we use the conservation laws encoded in the Killing vector $\boldsymbol{\xi}$ (i.e. that $\boldsymbol{u} \cdot \boldsymbol{\xi}$ is conserved along a geodesic). Our strategy is therefore to incorporate the energy Killing vector $\boldsymbol{\xi}$ into the description of the stationary observer's velocity.

With the Schwarzschild metric component $g_{t t}$, this gives a velocity com-
${ }^{8}$ Recall that $\xi$ has components $\xi^{\mu}=$ ( $1,0,0,0$ ).
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-090.jpg?height=355&width=377&top_left_y=752&top_left_x=319)

Fig. 22.2 A plot of $\omega_{R}$ against $R$ for $R>2 M$.

We can summarize the important results that we shall use to compute trajectories:
- For massive particles, the particle velocity $\boldsymbol{u}$ is tangent to the world line and has the property $\boldsymbol{u} \cdot \boldsymbol{u}=-1$.
■ For photons we have $\boldsymbol{u} \cdot \boldsymbol{u}=0$. The interval $\mathrm{d} s=0$ on the world line provides a useful constraint.
- We compute conservation laws using the Killing rule, that says that if the metric components are independent of variable $x^{\alpha}$, the component $u_{\alpha}$ is conserved along a geodesic.
To access the coordinate velocity $\mathrm{d} x^{i} / \mathrm{d} t$ we use the trick
$$
\begin{equation*}
\frac{\mathrm{d} x^{i}}{\mathrm{~d} t}=\frac{\mathrm{d} x^{i}}{\mathrm{~d} \tau} \cdot \frac{\mathrm{~d} \tau}{\mathrm{~d} t}=\frac{1}{u^{0}} \cdot \frac{\mathrm{~d} x^{i}}{\mathrm{~d} \tau} \tag{22.17}
\end{equation*}
$$
${ }^{9}$ The approach described here is covered in many books, but beware that conventions for defining some of the terms differ. We follow the approach and notation used in Hartle.
ponent at a radius $r$ of
$$
\begin{equation*}
u_{\mathrm{obs}}^{t}(r)=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \tag{22.14}
\end{equation*}
$$
or in terms of one of the Killing vectors, ${ }^{8} \boldsymbol{u}_{\text {obs }}(r)=(1-2 M / r)^{-\frac{1}{2}} \boldsymbol{\xi}$, which is the key equation. This expression means we can write eqn 22.12 in terms of the Killing vector $\boldsymbol{\xi}$. Specifically, the energy measured by the stationary observer sat at a radius $R$ is
$$
\begin{equation*}
\hbar \omega_{R}=\left(1-\frac{2 M}{R}\right)^{-\frac{1}{2}}(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{R} \tag{22.15}
\end{equation*}
$$
where subscript $R$ implies the quantity is evaluated at this radius. For the energy measured by the observer at infinity, the factor $(1-2 M / r)^{-\frac{1}{2}} \rightarrow 1$ and so $\hbar \omega_{\infty}=(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{\infty}$. However, since the photon is moving along a geodesic, the quantity $\boldsymbol{\xi} \cdot \boldsymbol{p}$ is conserved in the motion, and so $(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{R}=(-\boldsymbol{\xi} \cdot \boldsymbol{p})_{\infty}$ and we must therefore have
$$
\begin{equation*}
\omega_{\infty}=\omega_{R}\left(1-\frac{2 M}{R}\right)^{\frac{1}{2}} \tag{22.16}
\end{equation*}
$$

The frequency at infinity is less than the frequency at the point $R$ (see Fig. 22.2). One can rationalize this result by saying that the photon's energy is lowered through its climbing of the gravitational potential.

\subsection*{22.3 Motion in Schwarzschild spacetime}

We now turn to the motion of massive particles in the Schwarzschild geometry. ${ }^{9}$ It might be assumed that we need to solve the full geodesic equation to understand the motion of particles and photons. Fortunately for us, the only ingredients needed are the constants of the motion given by the Killing vectors, and the velocity identity $\boldsymbol{u} \cdot \boldsymbol{u}=-1$.

\section*{Example 22.1}

For simplicity, we assume motion in the equatorial plane of the geometry, which is to say that we fix $\theta=\pi / 2$ and so $u^{\theta}=0$. To analyse motion of a particle we start by using the velocity identity for the particle $\boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} u^{\mu} u^{\nu}=-1$, to write
$$
\begin{equation*}
-\left(1-\frac{2 M}{r}\right)\left(u^{t}\right)^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(u^{r}\right)^{2}+r^{2}\left(u^{\phi}\right)^{2}=-1 \tag{22.18}
\end{equation*}
$$

This can be rewritten using the conserved quantities $\tilde{E}$ and $\tilde{L}$ as
$$
\begin{equation*}
-\left(1-\frac{2 M}{r}\right)^{-1} \tilde{E}^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+\frac{\tilde{L}^{2}}{r^{2}}=-1 \tag{22.19}
\end{equation*}
$$

A little massaging of the above expression gives
$$
\begin{equation*}
\frac{\tilde{E}^{2}-1}{2}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+\frac{1}{2}\left[\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r^{2}}\right)-1\right] \tag{22.20}
\end{equation*}
$$

Now we introduce a new constant energy-like variable, ${ }^{10} \mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2$, and define an effective potential for motion in the Schwarzschild geometry of
$$
\begin{equation*}
V_{\mathrm{eff}}(r)=\frac{1}{2}\left[\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r^{2}}\right)-1\right]=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{M \tilde{L}^{2}}{r^{3}} \tag{22.21}
\end{equation*}
$$
and we end up with the familiar equation $\mathcal{E}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\text {eff }}(r)$.

The result of the last example is that the motion of a massive particle obeys the effective-potential equation
$$
\begin{equation*}
\mathcal{E}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\mathrm{eff}}(r) \tag{22.22}
\end{equation*}
$$
with $\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2$ and
$$
\begin{equation*}
V_{\mathrm{eff}}(r)=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{M \tilde{L}^{2}}{r^{3}} \tag{22.23}
\end{equation*}
$$

This is similar to the equation for motion in an effective potential that we saw ${ }^{11}$ in Chapter 20 for Newtonian motion. We conclude that motion takes place in an effective potential $V_{\text {eff }}$. The potential for the Schwarzschild geometry is shown in Fig. 22.3. Compared to the Newtonian potential, this one has more structure: specifically, with increasing $r$, an initial increase in $V_{\text {eff }}$ to a maximum, followed by behaviour that looks similar to its Newtonian cousin. This new structure leads to the richness of the new trajectories allowed by relativity.

Example 22.2
We can restore factors of $c$ and $G$ to find
$$
\begin{equation*}
V_{\mathrm{eff}}=\frac{1}{c^{2}}\left(-\frac{G M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{G M \tilde{L}^{2}}{c^{2} r^{3}}\right) \tag{22.24}
\end{equation*}
$$

Now define the non-relativistic, Newtonian energy $E_{\mathrm{N}}$ via
$$
\begin{equation*}
\tilde{E}=\frac{m c^{2}+E_{\mathrm{N}}}{m c^{2}} \tag{22.25}
\end{equation*}
$$

This allows us to write
$$
\begin{equation*}
E_{\mathrm{N}}=\frac{m}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+\frac{L^{2}}{2 m r^{2}}-\frac{G M m}{r}-\frac{G M L^{2}}{m c^{2} r^{3}} \tag{22.26}
\end{equation*}
$$
where $L=m \tilde{L}$. We conclude that the usual Newtonian energy equation is augmented at order $1 / c^{2}$ by the relativistic, $1 / r^{3}$ term.

We are now in the position to be able to provide a description of the motion. If we are interested in what would be measured by an observer, we will need access to the vielbein components for the Schwarzschild geometry in order to shift into the orthonormal frame of the observer. The vielbein for a stationary observer is given in the margin. It is worth noting that for this observer at infinity, measurements of the $\hat{t}$ and $\hat{r}$ components of a vector will give the coordinate values as both of the relevant vielbein components are unity at infinity.
${ }^{10}$ Particles coming in from infinity have $\tilde{E}>1$ and so have effective energy $\mathcal{E}>$ E.
0.
${ }^{11}$ The equations of motion differ by a factor of $m$. The Newtonian version has factor of $m$. The Newtonian version has
an effective potential energy $U_{\text {eff }}$ and so an effective potential
an effective potential
$$
V_{\mathrm{eff}}(r)=\frac{U_{\mathrm{eff}}(r)}{m}=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}
$$

The relativistic version has an extra term with a $1 / r^{3}$ dependence.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-091.jpg?height=395&width=354&top_left_y=829&top_left_x=1226)

Fig. 22.3 The relativistic effective potential for a particle (solid line, with $\tilde{L} / M=4.3$ ), compared to the Newtonian one (dotted line).

The vielbein components for an observer in the orthonormal frame (where the observer is stationary with respect to the coordinate frame) has components
$$
\begin{aligned}
& \left(\boldsymbol{e}_{t}\right)^{\hat{t}}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \\
& \left(\boldsymbol{e}_{r}\right)^{\hat{r}}=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \\
& \left(e_{\theta}\right)^{\hat{\theta}}=r \\
& \left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}=r \sin \theta
\end{aligned}
$$

It is important to remember that if we want to shift into the frame of a moving observer we will need a different vielbein. An example we have seen before that we will use several times is the freely falling observer's frame.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-092.jpg?height=230&width=58&top_left_y=407&top_left_x=453)

Fig. 22.4 The radial plunge.
${ }^{12}$ At infinity, spacetime is flat and the particle is at rest. Initially, the components of the velocity must therefore be $u^{\mu}(r=\infty)=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=(1,0,0,0)$. Recall also that the energy
$$
\tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t}
$$
is constant along a geodesic. For $r \rightarrow$ $\infty$ we have $\tilde{E}=1$. Also recall that the (constant) effective energy is
$$
\mathcal{E}=\frac{\tilde{E}^{2}-1}{2}=0
$$
${ }^{13}$ There is no mechanism to pick up angular velocity, hence the zeros for the $\theta$ and $\phi$ components of velocity.
${ }^{14}$ The falling observer's local frame is flat, but moving with respect to the coordinate frame, so differs from the conventional orthonormal frame we often use. To find this frame, note that we always take $\boldsymbol{e}_{\hat{t}}=\boldsymbol{u}$. Then, for convenience choose $\left(\boldsymbol{e}_{\hat{\theta}}\right)^{\mu}=(0,0,1,0)$ and $\left(\boldsymbol{e}_{\hat{\phi}}\right)^{\mu}=(0,0,0,1)$. This means that the one remaining vielbein component is
$$
\left(\boldsymbol{e}_{\hat{r}}\right)^{\mu}=\left(\begin{array}{c}
\frac{-\sqrt{2 M / r}}{1-2 M / r}  \tag{22.30}\\
1 \\
0 \\
0
\end{array}\right)
$$
${ }^{15}$ In this section, it is necessary to fix the value of $r$ at some value of the proper time $\tau$ in order to fix the integration constant.
${ }^{16}$ A useful quantity to note for computations is the coordinate velocity
$$
\begin{equation*}
\frac{\mathrm{d} r}{\mathrm{~d} t}=-\left(1-\frac{2 M}{r}\right)\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{22.34}
\end{equation*}
$$

The magnitude of this quantity is the coordinate escape velocity of a particle, as discussed in the exercises.

\subsection*{22.4 Example: the radial plunge}

Let's consider a massive particle that starts at rest at infinity and plunges towards a star along a radial line (Fig. 22.4). This line is a geodesic.

\section*{Example 22.3}

We have immediately that the angular momentum $\tilde{L}=0$ since the plunge is radial. 'At rest at infinity' implies that at a great distance from the gravitating object we have ${ }^{12} u^{t}=\mathrm{d} t / \mathrm{d} \tau=1$ and so, for the duration of the freefall, $\tilde{E}=1$ and $\mathcal{E}=0$. The fact that $\tilde{E}=1$ allows us to pick out, at later times, that
$$
\begin{equation*}
u^{t}=\frac{\mathrm{d} t}{\mathrm{~d} \tau}=\left(1-\frac{2 M}{r}\right)^{-1} \tag{22.27}
\end{equation*}
$$

The effective energy $\mathcal{E}$ in eqn 22.22 for a radial plunge with $\tilde{L}=0$ and $\mathcal{E}=0$ gives us an equation of motion
$$
\begin{equation*}
0=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}-\frac{M}{r} \tag{22.28}
\end{equation*}
$$
which provides us with an expression for the radial velocity $u^{r}=\mathrm{d} r / \mathrm{d} \tau=$ $-(2 M / r)^{\frac{1}{2}}$, where the sign is chosen so that the particle is falling towards the gravitating star. We conclude that the velocity during the fall is given by a vector $\boldsymbol{u}$ with components in the coordinate frame of ${ }^{13}$
$$
\begin{equation*}
u^{\mu}=\left(\left(1-\frac{2 M}{r}\right)^{-1},-\left(\frac{2 M}{r}\right)^{\frac{1}{2}}, 0,0\right) \tag{22.29}
\end{equation*}
$$

The vielbein components for this observer's frame can also be computed. ${ }^{14}$

Let's consider how much proper time and coordinate time elapses for the falling particle.

\section*{Example 22.4}

First the proper time. We rewrite eqn 22.28 as an integral
$$
\begin{equation*}
\int_{r}^{0} r^{\frac{1}{2}} \mathrm{~d} r=-\int_{\tau}^{\tau_{*}}(2 M)^{\frac{1}{2}} \mathrm{~d} \tau \tag{22.31}
\end{equation*}
$$
where the negative square root has been chosen for an inward-travelling particle and $\tau_{*}$ fixes the value of the proper time ${ }^{15}$ at the end of the plunge, when $r=0$. This can be integrated with the result that
$$
\begin{equation*}
r(\tau)=\left(\frac{3}{2}\right)^{\frac{2}{3}}(2 M)^{\frac{1}{3}}\left(\tau_{*}-\tau\right)^{\frac{2}{3}} \tag{22.32}
\end{equation*}
$$

We can then say that the proper time for a falling astronaut is given by
$$
\begin{equation*}
\tau(r)=\tau_{*}-\left(\frac{2}{3}\right)\left(\frac{1}{2 M}\right)^{\frac{1}{2}} r^{\frac{3}{2}} \tag{22.33}
\end{equation*}
$$

The variation of $\tau$ with the Schwarzschild radius coordinate $r$ is shown in Fig. 22.5. Notice that, from an arbitrary starting value of $r$ it takes a finite amount of proper time to reach any value of $r$, as we might expect.

Next, the amount of elapsed coordinate time $t$. Since $\left(e_{t}\right)^{\bar{t}}=1$ at infinity, this is also the time measured by an observer at infinity watching the particle plunge. We have ${ }^{16}$
$$
\begin{equation*}
\frac{\mathrm{d} t}{\mathrm{~d} r}=\frac{\mathrm{d} t}{\mathrm{~d} \tau} \frac{\mathrm{~d} \tau}{\mathrm{~d} r}=\frac{u^{t}}{u^{r}}=-\left(\frac{2 M}{r}\right)^{-\frac{1}{2}}\left(1-\frac{2 M}{r}\right)^{-1} \tag{22.35}
\end{equation*}
$$

This equation can be integrated to give
$$
\begin{equation*}
t=t_{*}+2 M\left[-\frac{2}{3}\left(\frac{r}{2 M}\right)^{\frac{3}{2}}-2\left(\frac{r}{2 M}\right)^{\frac{1}{2}}+\ln \left|\frac{\left(\frac{r}{2 M}\right)^{\frac{1}{2}}+1}{\left(\frac{r}{2 M}\right)^{\frac{1}{2}}-1}\right|\right] \tag{22.36}
\end{equation*}
$$
where $t_{*}$ gives the time coordinate when $r=0$. This function is graphed in Fig. 22.6.

There are some curious facts about the results from the last example: the most notable being that, from the point of view of the observer at infinity, it takes an infinite amount of coordinate time for an in-falling astronaut to reach $r=2 M$ (but a finite proper time). We will return to this point in Chapter 25.
Finally, let's examine the coordinate velocity of the particle that plunges radially in a Schwarzschild metric.

\section*{Example 22.5}

We won't assume the particle starts from rest this time. The velocity identity gives us
$$
\begin{align*}
-1 & =\boldsymbol{u} \cdot \boldsymbol{u} \\
& =-\left(1-\frac{2 M}{r}\right)\left(u^{t}\right)^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(u^{r}\right)^{2} \\
& =\left[-\left(1-\frac{2 M}{r}\right)+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}\right]\left(u^{t}\right)^{2} \\
& =\left[-\left(1-\frac{2 M}{r}\right)+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}\right]\left(u_{t}\right)^{2}\left(1-\frac{2 M}{r}\right)^{-2}, \tag{22.37}
\end{align*}
$$
where, in the final line we've written the equation in terms of the constant of the motion $\boldsymbol{u} \cdot \boldsymbol{\xi}=u_{t}$. Solving for $(\mathrm{d} r / \mathrm{d} t)^{2}$ we find
$$
\begin{equation*}
\left(\frac{\mathrm{d} r}{\mathrm{~d} t}\right)^{2}=\left(1-\frac{2 M}{r}\right)^{2}\left[1-\left(1-\frac{2 M}{r}\right) \frac{1}{\left(u_{t}\right)^{2}}\right] \tag{22.38}
\end{equation*}
$$

This coordinate velocity vanishes as the particle approaches $r=2 M$ (Fig. 22.7).
What does a stationary observer determine as the coordinate velocity? We won't assume that they are at infinity here and so we shift to the orthonormal frame using the vielbein. Remember that for 1-forms like $\boldsymbol{d} t$ and $\boldsymbol{d} r$ we have $\boldsymbol{\omega}^{\hat{\mu}}=\left(\boldsymbol{e}_{\mu}\right)^{\hat{\mu}} \boldsymbol{\omega}^{\mu}$. We find
$$
\begin{align*}
& \boldsymbol{d} \hat{t}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \boldsymbol{d} t \\
& \boldsymbol{d} \hat{r}=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \boldsymbol{d} r . \tag{22.39}
\end{align*}
$$

Noting that the derivative $\mathrm{d} r / \mathrm{d} t \equiv \boldsymbol{d} r / \boldsymbol{d} t$, we have that the stationary observer observes a plunge with a velocity
$$
\begin{equation*}
\frac{\mathrm{d} \hat{r}}{\mathrm{~d} \hat{t}}=\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} r}{\mathrm{~d} t} \tag{22.40}
\end{equation*}
$$

Irrespective of $u_{t}$, and hence the initial velocity, this locally measured velocity approaches unity or, in real units the speed of light, as $r \rightarrow 2 M$ (Fig. 22.7).

Using only the velocity identity and the conservation of energy and angular momentum, we have shown a number of curious relativistic effects in a particle's motion. We'll pick up these points in a few chapters' time. In the next chapter, we turn to orbits.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-093.jpg?height=335&width=309&top_left_y=394&top_left_x=1223)

Fig. 22.5 The evolution of the coordinate $r$ with the proper time $\tau$ for the radial plunge.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-093.jpg?height=363&width=323&top_left_y=885&top_left_x=1219)

Fig. 22.6 The evolution of the coordinate $r$ with the coordinate time $t$ for the radial plunge. The dotted line shows $r_{\mathrm{S}}=2 M$.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-093.jpg?height=366&width=326&top_left_y=1368&top_left_x=1209)

Fig. 22.7 Example evolution of the coordinate velocity $v=\mathrm{d} r / \mathrm{d} t$ with radial coordinate $r$ for the radial plunge (solid line). The dashed line shows the coordinate velocity $\hat{v}=\mathrm{d} \hat{r} / \mathrm{d} \hat{t}$ measured by a stationary observer.

\section*{Chapter summary}
- Motion in the Schwarzschild geometry can often be computed using the velocity $\boldsymbol{u}$, subject to the constraint that $\boldsymbol{u} \cdot \boldsymbol{u}=-1$. The constants of the motion $\tilde{E}$ and $\tilde{L}$ follow from the independence of the metric components from coordinates $t$ and $\phi$.
- The effective potential method allows motion to be analysed in terms of a relativistic $V_{\text {eff }}$.
- A particle that plunges from rest at infinity has $\mathcal{E}=0$ and a velocity $\boldsymbol{u}$ with components
$$
\begin{equation*}
u^{\mu}=\left(\left(1-\frac{2 M}{r}\right)^{-1},-\left(\frac{2 M}{r}\right)^{\frac{1}{2}}, 0,0\right) \tag{22.41}
\end{equation*}
$$

\section*{Exercises}
22.1) Consider the acceleration $\boldsymbol{a}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}$. By the geodesic equation, the components of the acceleration $a^{\mu}=u^{\alpha} u^{\mu}{ }_{; \alpha}$ vanish in free fall.
(a) What are the components of the acceleration for a particle at rest in a static gravitational field with metric components $g_{\mu \nu}$ ?
(b) What are the components of the acceleration of a particle at rest at a fixed value of $r$, in a Schwarzschild coordinate system.
(c) What is the proper acceleration $\alpha$ (i.e. the acceleration measured in the frame of the particle)?
(22.2) Although we have used a variety of shortcuts to compute the equations of motion, the same information is available from the geodesic equation.
(a) Compute the components of the geodesic equation for the Schwarzschild geometry with line element given in the form of eqn 22.2 .
If you've done this previously for the line element expressed in different variables, you could simply re-express the equations in terms of $(t, r, \theta, \phi)$.
(b) How do these expressions simplify for motion in the plane with $\theta=\pi / 2$ ?
(c) Show that the equation of motion for the $r$ coordinate is consistent with eqn 22.22 .
(22.3) Derive eqn 22.16 using the vielbein components, and the fact that $p_{t}$ is conserved on the geodesic.
(22.4) Verify eqn 22.26 in the Newtonian limit.
22.5) A stationary observer on the surface of a spherical mass $M$ of radius $R$ launches a projectile at its escape velocity. It does not experience any force during its motion and so it follows a geodesic.
(a) What is the escape velocity of the projectile measured in the observer's rest frame?
(b) What is the energy of the projectile as measured by the observer?
(22.6) Gravitational clock effect: Consider the Schwarzschild geometry describing the gravitational field of the Earth, which is rotating on its axis with angular velocity $\omega$.
(a) For an observer at rest on the Earth's surface at the equator, show that the proper time interval is given approximately by
$$
\begin{equation*}
\mathrm{d} \tau_{1} \approx\left(1-\frac{M}{r}-\frac{r^{2} \omega^{2}}{2}\right) \mathrm{d} t \tag{22.42}
\end{equation*}
$$
(b) Now consider a second observer who flies eastwards around the equator with velocity $v$ at a height $h$, whose measures a proper time interval $\mathrm{d} \tau_{2}$. Show that
$$
\begin{equation*}
\mathrm{d} \tau_{1}-\mathrm{d} \tau_{2} \approx\left[-\frac{M h}{r^{2}}+\frac{(2 r \omega+v) v}{2}\right] \mathrm{d} t \tag{22.43}
\end{equation*}
$$
(c) Define $\Delta=\left(\mathrm{d} \tau_{1}-\mathrm{d} \tau_{2}\right) / \mathrm{d} \tau_{1}$ and estimate this quantity for an eastward flight around the Earth.
(d) Constrast the value of $\Delta$ for a similar flight in a westward direction.
This experiment was carried out in the early 1970s by Hafele and Keating, whose results were consistent with predictions in this problem fwhich follows the method in Ryder (2009)].
(22.7) Two observers start on the same radial line in the Schwarzschild geometry, one at rest at $r_{1}$ and an-
other at rest at $r_{2}>r_{1}$. At $t=0$, the observer at $r_{2}$ begins to fall freely. What is the relative velocity of the observers as they meet?
Hint: Recall from Chapter 2 that $\gamma$, corresponding to the instantaneous relative velocity $v_{\text {rel }}$, can be found by taking the dot product of the two observers' velocity vectors: $-\gamma\left(v_{\text {rel }}\right)=\boldsymbol{u} \cdot \boldsymbol{v}$.

\section*{23}

\section*{Orbits in the Schwarzschild geometry}
23.1 Orbits for massive particles
23.2 Stable circular orbits 248 23.3 Precession of the perihelion
Chapter summary 252

Exercises

The Schwarzschild line element is given by
$$
\begin{aligned}
\mathrm{d} s^{2}= & -\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2} \\
& +\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2} \\
& +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)
\end{aligned}
$$

The conserved quantities are
$$
\tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t}
$$
and
$\tilde{L}=r^{2} \sin ^{2} \theta u^{\phi}$.
${ }^{1}$ See Exercise 22.2.

I am about to make my last voyage, a great leap in the dark Thomas Hobbes (1588-1679)

The description of orbits given in Chapter 20 proved a crowning triumph of Newtonian mechanics. One of the first successes of general relativity was to provide predictions of orbits that, although similar to Kepler and Newton's ellipses, actually provided closer agreement with observation. We now turn to the possible orbits in the Schwarzschild geometry. These are a special class of geodesic, where a particle executes periodic motion by freely falling along a closed path in space. As in the last chapter, we will make use of the Schwarzschild line element and the conserved quantities $E$ and $L$. Previously, we have confined our attention to those plunging motions with $L=0$. We now relax that constraint in order to allow angular motion. Our discussion of the resulting stable orbits will yield the famous prediction of the precession of the perihelion of an orbit in the Schwarzschild geometry.

\subsection*{23.1 Orbits for massive particles}

Orbits made by massive particles are on geodesics parametrized by their proper time $\tau$. For simplicity, we restrict our attention to particles moving in an equatorial plane, so that we can set $\theta=\pi / 2$ and so $u^{\theta}=$ $\mathrm{d} \theta / \mathrm{d} \tau=0$. There is no loss of generality here since, as in the Newtonian case, orbiting particles are confined to a plane, as we shall now prove.

Example 23.1
If we consider the geodesic equation found by varying the Schwarzschild line element with respect to the angle ${ }^{1} \theta$, we obtain
$$
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{~d} \tau}\left(r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} \tau}\right)=r^{2} \sin \theta \cos \theta\left(\frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}\right)^{2} \tag{23.1}
\end{equation*}
$$

This equation of motion is solved by $\theta=\pi / 2$ for all $\tau$. At this angle the right-hand side of the equation is zero and so the quantity $r^{2} \mathrm{~d} \theta / \mathrm{d} \tau$ is conserved. Since we initially fix $\frac{d}{d \tau} \theta=0$, the particle never acquires any acceleration, and the motion is confined to the plane. Just as in the Newtonian case, the confinement to the plane can be attributed to conservation of angular momentum. That is, if we rotate our coordinates such that we initially have $\theta=\frac{\pi}{2}$, then the equation of motion for $u^{\phi}$ gives $\frac{\mathrm{d}}{\mathrm{d} \tau} \tilde{L}=0$ on the whole geodesic.

Our strategy will be to understand the possible orbits by analysing an effective potential, much as we did both in the Newtonian case and also in the last chapter. Using the effective energy variable $\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2$,
$$
\begin{equation*}
\mathcal{E}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\mathrm{eff}}(r) \tag{23.2}
\end{equation*}
$$

Using the results from Example 22.1 we have an effective potential for particles in the Schwarzschild geometry, given by
$$
\begin{equation*}
V_{\mathrm{eff}}(r)=-\frac{M}{r}+\frac{\tilde{L}^{2}}{2 r^{2}}-\frac{M \tilde{L}^{2}}{r^{3}} \tag{23.3}
\end{equation*}
$$

As we noted in the last chapter, this setup is designed to look as much like the Newtonian version as possible. ${ }^{2}$ Here, the extra attractive term $-M \tilde{L}^{2} / r^{3}$ is the source of the richness of relativistic trajectories. This term is largest for large $M$ and $\tilde{L}$ and small values of $r$.

An example of the effective potential $V_{\text {eff }}$ is shown in Fig. 23.1, where we see its two characteristic extrema: (i) a minimum at a radius $r_{+}$, which resembles the minimum in the Newtonian effective potential; and (ii) a maximum is a smaller radius $r_{-}$, which has no analogue in Newtonian physics. These turning points in the relativistic potential, occurring for $V_{\text {eff }}^{\prime}\left(r_{ \pm}\right)=0$, are found at
$$
\begin{equation*}
r_{ \pm}=\frac{\tilde{L}^{2}}{2 M}\left[1 \pm\left(1-\frac{12 M^{2}}{\tilde{L}^{2}}\right)^{\frac{1}{2}}\right] \tag{23.4}
\end{equation*}
$$

This equation implies that the nature of the extrema depends on the ratio $\tilde{L} / M$. That is, a combination of the angular momentum of the particle in motion and the mass of the gravitating object determines the effective potential, just as in Newtonian physics.

\section*{Example 23.2}

In Fig. 23.2, we show some examples of motion in the potential determined by one choice of $M$ and (non-zero) $\tilde{L}$. In these plots, we represent $V_{\text {eff }}(r)$ and the particle's value of $\mathcal{E}$. Where $\mathcal{E}=V_{\text {eff }}$ and the lines meet, the particle has no velocity in the radial direction, although it is important to note that the particle has a non-zero value of $\tilde{L}$ (and therefore $u^{\phi}$ ) and so is still in motion.
- Figure 23.2(a) shows a circular orbit, with a particle sat at the minimum of $V_{\text {eff }}$ at a (constant) radius $r_{+}$. The particle has effective energy $\mathcal{E}=V_{\text {eff }}\left(r_{+}\right)$, which takes a negative value, corresponding to the particle being bound in a potential.
- Figure 23.2(b) shows a particle in-falling from infinity (so therefore having a positive effective energy ${ }^{3} \mathcal{E}$ ), but colliding with the peak in the potential, which scatters it back off to infinity.
- Figure 23.2(c) shows the motion of a particle with a still higher energy. This one - Figure 23.2 (c) shows the motion of a particle with a still higher energy. This one
is not scattered since its effective energy is too high for it to hit the peak in the potential $V_{\text {eff }}\left(r_{-}\right)$. Instead, this particle spirals towards the centre of the gravitating mass, where it presumable crashes into the mass distribution at some small radius. This trajectory is a consequence of the relativistic potential being distinct from the Newtonian version, in that the latter has an energetic barrier at small $r$. This means that a Newtonian particle with positive energy, as long as it has some angular momentum, will always be deflected by the potential (assuming it doesn't crash into the source of the potential first!).
${ }^{2}$ Recall the Newtonian effective potential energy is given by
$$
U_{\mathrm{eff}}(r)=\frac{L^{2}}{2 m r^{2}}-\frac{G M m}{r}
$$
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-097.jpg?height=391&width=372&top_left_y=563&top_left_x=1211)

Fig. 23.1 The relativistic effective potential for given values of $\tilde{L} / M$.
(a) $V_{\text {eff }}(\vec{r})$
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-097.jpg?height=433&width=383&top_left_y=1033&top_left_x=1209)
(c) $V_{\text {eff }}(\vec{r})$
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-097.jpg?height=212&width=223&top_left_y=1477&top_left_x=1220)

Fig. 23.2 Some allowed trajectories in the relativistic potential (computed using eqn 23.15). (a) Circular; (b) an encounter with the star leading to scattering; (c) a spiral into the star.
${ }^{3}$ Recall that particles arriving from infinity have $\tilde{E}>1$ so $\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2>$
${ }^{4}$ These particles will simply spiral into the origin.
${ }^{5}$ In addition to the minimum, it's notable that the maximum of $V_{\text {eff }}$ also represents a possible circular orbit. Being a maximum, it is manifestly unstable. The analogous trajectory will be considered for photons in the next chapter.

One thing that we can spot from eqn 23.4 and Fig. 23.1 is that if $\tilde{L} / M<$ $\sqrt{12}(=3.464 \ldots)$ then there is only one extremum, a point of inflection. For the corresponding motions at low $L / M$ there is not enough angular momentum to stabilize an orbit. The stable orbits are only possible for $\tilde{L} / M \geq \sqrt{12}$ and, as we shall discover below, these stable orbits are almost, but not quite, elliptical. The difference between these orbits and their Newtonian analogues is that the motion can precess, such that the near-ellipses traced by the trajectories rotate.
It is also the case that if $\tilde{L} / M=4$, then the maximum of $V_{\text {eff }}$ occurs at $r_{-}=4 M$ where $V_{\text {eff }}=0$, which is significant as it means that particles coming in from infinity, which have $\mathcal{E}>0$, can never be scattered by such a potential. ${ }^{4}$ If the angular momentum is greater than $4 M$ then there are possible trajectories that can scatter particles coming in from infinity. Less angular momentum than this implies a spiral into the origin for $\mathcal{E}>0$ particles.

\subsection*{23.2 Stable circular orbits}

In the Newtonian case, we saw how the minimum of the effective potential gave us the circular orbits. We can repeat the computation for the relativistic case and show the properties of circular orbits here too. ${ }^{5}$ Stable circular orbits occur when $r=r_{+}$in eqn 23.4. These depend on $\tilde{L} / M$, with the minimum angular momentum occurring for $\tilde{L} / M=\sqrt{12}$, for which we have $r_{+}=6 \mathrm{M}$. This represents a minimum radius for stable circular orbits in the Schwarzschild geometry.

Example 23.3
From Chapter 20 we know that for a circular orbit we have that the effective potential is a minimum and that the effective energy $\mathcal{E}=\left(\tilde{E}^{2}-1\right) / 2$ equals the value of the effective potential $V_{\text {eff }}\left(r_{+}\right)$. Rearranging eqn 23.2 , this amounts to the condition
$$
\begin{equation*}
\tilde{E}^{2}=\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r_{+}^{2}}\right) \tag{23.5}
\end{equation*}
$$

This equation, along with eqn 23.4 , can be usefully combined to find the ratio $\tilde{L} / \tilde{E}$ for a circular orbit. From eqn 23.4 we have
$$
\begin{equation*}
\left(2 M r_{+}-\tilde{L}^{2}\right)^{2}=\tilde{L}^{4}\left(1-\frac{2 M^{2}}{\tilde{L}^{2}}\right) \tag{23.6}
\end{equation*}
$$
from which, on rearranging, we obtain
$$
\begin{equation*}
\frac{\bar{L}^{2}}{r_{+}^{2}}=\frac{M}{r_{+}-3 M} \tag{23.7}
\end{equation*}
$$

Combining with eqn 23.5, we find
$$
\begin{equation*}
\tilde{E}^{2}=\left(1-\frac{2 M}{r_{+}}\right)\left(\frac{r_{+}-2 M}{r_{+}-3 M}\right) \tag{23.8}
\end{equation*}
$$
from which we can write
$$
\begin{equation*}
\left(\frac{\tilde{L}}{\tilde{E}}\right)^{2}=M r_{+}^{2}\left(1-\frac{2 M}{r_{+}}\right)^{-2} \tag{23.9}
\end{equation*}
$$

The result of the previous example is useful, since it shows that the angular velocity of a particle in an equatorial orbit is given by ${ }^{6}$
$$
\begin{equation*}
\frac{\mathrm{d} \phi}{\mathrm{~d} t}=\frac{\mathrm{d} \phi / \mathrm{d} \tau}{\mathrm{~d} t / \mathrm{d} \tau}=\frac{1}{r^{2}}\left(1-\frac{2 M}{r}\right)\left(\frac{\tilde{L}}{\tilde{E}}\right) \tag{23.10}
\end{equation*}
$$

Plugging the result for $\tilde{L} / \tilde{E}$ into eqn 23.10 leads to relativistic version of Kepler's law given by
$$
\begin{equation*}
\left(\frac{\mathrm{d} \phi}{\mathrm{~d} t}\right)^{2}=\frac{M}{r^{3}} \tag{23.11}
\end{equation*}
$$

This is, of course, the same as Kepler's original version of his law, and so we conclude that Kepler's third law for circular orbits is not altered by relativity.

\subsection*{23.3 Precession of the perihelion}

The power of general relativity lies partially in explaining effects we find in observations that could not occur in Newtonian gravity. Historically, one of the most important was the precession of the perihelion of mercury. This is one of the three classical solar-system tests of general relativity. Below we shall derive the equation for the trajectory of a particle in the relativistic effective potential. The general idea is illustrated in the following example.

\section*{Example 23.4}

The precessing orbit is shown in Fig. 23.3 (for a vastly exaggerated angle of precession compared to what we will compute below). The particle oscillates between two points on the effective potential curve, hitting maximum and minimum distances from the origin. As it does so, the trajectories resemble ellipses whose axes precess in the plane of motion.

Since our trajectories are confined to the plane $\theta=\pi / 2$, then in order to find the shape of planetary orbits we must write an equation for the coordinate $r$ in terms of $\phi$ (or vice-versa). Since we are not solving the full geodesic equations, but instead exploiting (i) the square of $\boldsymbol{u}$ and (ii) the conserved quantities $\tilde{E}$ and $\tilde{L}$, the way to do this is to find $\mathrm{d} r / \mathrm{d} \tau$ and $\mathrm{d} \phi / \mathrm{d} \tau$ and take their ratio to find $\mathrm{d} r / \mathrm{d} \phi$.
Starting with the velocity identity $\boldsymbol{u} \cdot \boldsymbol{u}=-1$, for fixed $\theta$ we have
$$
\begin{equation*}
g^{t t}\left(u_{t}\right)^{2}+g^{\phi \phi}\left(u_{\phi}\right)^{2}+g^{r r}\left(u_{r}\right)^{2}=-1 \tag{23.12}
\end{equation*}
$$

Noting that $g^{r r}\left(u_{r}\right)^{2}=g_{r r}\left(u^{r}\right)^{2}$ for this diagonal metric, ${ }^{7}$ we have
$$
\begin{equation*}
\left(u^{r}\right)^{2}=\left(\frac{\mathrm{d} r}{\mathrm{~d} \tau}\right)^{2}=\frac{-1-g^{t t} \tilde{E}^{2}-g^{\phi \phi} \tilde{L}^{2}}{g_{r r}} \tag{23.13}
\end{equation*}
$$
${ }^{6}$ We make use of
$$
u^{\mu}=\left(\frac{\mathrm{d} t}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \tau}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \theta}{\mathrm{~d} \tau}, \frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}\right)
$$
and the equations for conserved quantities
$$
\begin{gathered}
\tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t} \\
\tilde{L}=r^{2} u^{\phi}
\end{gathered}
$$

The three classical solar system tests of relativity are:
- Precession of the perihelion of mercury;
- Bending of light by the sun;
- Gravitational redshift.

We met gravitational redshift in Chapter 6 . In Chapter 9, we also met a fourth test: Shapiro time delay.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-099.jpg?height=212&width=368&top_left_y=1160&top_left_x=1219)

Fig. 23.3 The precessing orbit in the relativistic potential calculated using eqn 23.15. The motion is bounded at the points shown on the left, represented by the dotted lines on the right.

The components of the inverse Schwarzschild metric are
$$
\begin{aligned}
g^{t t} & =-\left(1-\frac{2 M}{r}\right)^{-1} \\
g^{r r} & =\left(1-\frac{2 M}{r}\right) \\
g^{\theta \theta} & =r^{-2} \\
g^{\phi \phi} & =(r \sin \theta)^{-2}
\end{aligned}
$$
${ }^{7}$ That is, $g^{r r}\left(u_{r}\right)^{2}=g^{r r}\left(g_{r r} u^{r}\right)^{2}$ and $g_{r r} g^{r r}=1$.
and so
$$
\begin{equation*}
\frac{\mathrm{d} r}{\mathrm{~d} \tau}= \pm\left\{\left[-1+\tilde{E}^{2}\left(1-\frac{2 M}{r}\right)^{-1}-\frac{\tilde{L}^{2}}{r^{2}}\right]\left(1-\frac{2 M}{r}\right)\right\}^{\frac{1}{2}} . \tag{23.14}
\end{equation*}
$$

Since we also know from $u^{\phi}=g^{\phi \phi} \tilde{L}$ that $\frac{\mathrm{d} \phi}{\mathrm{d} \tau}=\frac{\tilde{L}}{r^{2}}$, we are able to isolate the equation that gives us the trajectories in the Schwarzschild geometry through the multiplication $\frac{\mathrm{d} r}{\mathrm{~d} \tau} \frac{\mathrm{~d} \tau}{\mathrm{~d} \phi}$, which gives
$$
\begin{equation*}
\frac{\mathrm{d} r}{\mathrm{~d} \phi}= \pm \frac{r^{2}}{\tilde{L}}\left\{\left[-1+\tilde{E}^{2}\left(1-\frac{2 M}{r}\right)^{-1}-\frac{\tilde{L}^{2}}{r^{2}}\right]\left(1-\frac{2 M}{r}\right)\right\}^{\frac{1}{2}} . \tag{23.15}
\end{equation*}
$$

This is our basic equation that tells us the paths in Schwarzschild coordinates that particles can take. ${ }^{8}$ We can analyse this equation using some of the same techniques that we used in Chapter 20, where we introduced the variable ${ }^{9} u=1 / r$.
tion is
$\frac{\mathrm{d} r}{\mathrm{~d} \theta}=$
$$
\pm \frac{r^{2}}{L}\left\{2 m\left[E+\frac{G M m}{r}\right]-\frac{L^{2}}{r^{2}}\right\}_{(23.16)}^{\frac{1}{2}}
$$
or, using $u=1 / r$,
$\left(u^{\prime}\right)^{2}+u^{2}-\frac{2 G M m^{2}}{L^{2}} u-\frac{2 E m}{L^{2}}=0$.
Recall that we defined $u_{0}=G M m^{2} / L^{2}$ and $u_{0}\left(1-\epsilon^{2}\right)=-2 E m / L^{2}$ and wrote
$$
\left(u^{\prime}\right)^{2}+\left(u-u_{0}\right)^{2}-u_{0}^{2} \epsilon^{2}=0
$$
${ }^{9}$ Don't confuse $u=1 / r$ with the components of the velocity $\boldsymbol{u}$, despite their sharing a symbol.

\section*{Example 23.5}

Here we introduce the scaled variable $u=M / r$ and find
$$
\begin{equation*}
\left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}=\frac{\tilde{E}^{2}-(1-2 u)\left[1+(\tilde{L} / M)^{2} u^{2}\right]}{(\tilde{L} / M)^{2}} \tag{23.17}
\end{equation*}
$$

This equation can be rewritten in a form compatible with the Newtonian version. Specifically, it can be massaged into the expression
$$
\begin{equation*}
\left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}+\left(u-u_{0}\right)^{2}=6 u_{0}\left(u-u_{0}\right)^{2}+2\left(u-u_{0}\right)^{3}+\frac{M^{2}}{\tilde{L}^{2}}\left(\tilde{E}^{2}-\tilde{E}_{0}^{2}\right) \tag{23.18}
\end{equation*}
$$

Expanding out and comparing with eqn 23.17, we find that
$$
\begin{equation*}
6 u_{0}=1 \pm\left(1-\frac{12 M^{2}}{\tilde{L}^{2}}\right)^{\frac{1}{2}} \tag{23.19}
\end{equation*}
$$
and
$$
\begin{equation*}
E_{0}^{2}=\left(1-2 u_{0}\right)\left(1+\frac{\tilde{L}^{2}}{M^{2}} u_{0}^{2}\right) \tag{23.20}
\end{equation*}
$$

These expressions are consistent with the intuitions we built up about which orbits are possible. In order to compare more closely to the Newtonian version, we write $\epsilon u_{0}^{2}=\frac{M^{2}}{\tilde{L}^{2}}\left(\tilde{E}^{2}-\tilde{E}_{0}^{2}\right)$ and we have
$$
\begin{equation*}
\left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}+\left(u-u_{0}\right)^{2}-\epsilon^{2} u_{0}^{2}=6 u_{0}\left(u-u_{0}\right)^{2}+2\left(u-u_{0}\right)^{3} \tag{23.21}
\end{equation*}
$$

The terms on the right are relativistic corrections to the Newtonian expression on the left (see Sidenote 8). The Newtonian solution is $u=u_{0}(1+\epsilon \cos \phi)$. The first-order correction term is $6 u_{0}\left(u-u_{0}\right)^{2}$, which is of order $O\left(u_{0}^{3} \epsilon^{2}\right)=O\left(M^{3} \epsilon^{2} / r_{0}^{3}\right)$. In contrast, the second-order term is $2\left(u-u_{0}\right)^{3}$, which is of order $O\left(u_{0}^{3} \epsilon^{3}\right)=O\left(M^{3} \epsilon^{3} / r_{0}^{3}\right)$. For small eccentricities, the second-order correction can be ignored.

From the last example we conclude that the orbits are almost Newtonian ellipses with $u=u_{0}(1+\epsilon \cos \phi)$, but are slightly perturbed by the corrections terms. Considering only the larger, first-order corrections, we must solve
$$
\begin{equation*}
\left(u^{\prime}\right)^{2}=\left(1-6 u_{0}\right)\left(u-u_{0}\right)^{2}=u_{0}^{2} \epsilon^{2} \tag{23.22}
\end{equation*}
$$

To solve the equation, define $\psi=\left(1-6 u_{0}\right)^{\frac{1}{2}} \phi$ and $\mu=\left(u-u_{0}\right)$. In terms of these variables, we have an equation of motion
$$
\begin{equation*}
\left(\frac{\mathrm{d} u}{\mathrm{~d} \psi}\right)^{2}+\mu^{2}=\frac{u_{0}^{2} \epsilon^{2}}{\left(1-6 u_{0}\right)} \tag{23.23}
\end{equation*}
$$

The solution to this equation of motion is periodic in $\psi$. This means that for each orbit $\Delta \psi=2 \pi$. However, for the angle $\phi$ in which we are interested, $\Delta \phi \neq 2 \pi$. Instead, from the definition of $\psi$, we can infer that an orbit corresponds to a change in $\phi$ of
$$
\begin{equation*}
\Delta \phi=\frac{2 \pi}{\left(1-\frac{6 M}{r_{0}}\right)^{\frac{1}{2}}} \approx 2 \pi\left(1+\frac{3 M}{r_{0}}\right) \tag{23.24}
\end{equation*}
$$

The geometry for this effect is shown in Fig. 23.4. It implies that the change in the position of the perihelion can be written as $\Delta \phi=2 \pi+\delta \phi$ where for each orbit
$$
\begin{equation*}
\frac{\delta \phi}{2 \pi}=\frac{3 M}{r_{0}} . \tag{23.25}
\end{equation*}
$$

This provides the basis of our estimate for the precession of the perihelion of mercury.

Example 23.6
The orbit of Mercury has a period of 88 days $\left(\approx 7.6 \times 10^{6} \mathrm{~s}\right)$. If we take $r_{0}=$ $5.8 \times 10^{10} \mathrm{~m}$ and $M_{\odot}=2 \times 10^{30} \mathrm{~kg}$ then, restoring constants, we compute
$$
\begin{equation*}
\delta \phi=6 \pi \frac{G M_{\odot}}{c^{2} r_{0}} \approx 4.8 \times 10^{-7} \mathrm{rad} \text { per orbit. } \tag{23.26}
\end{equation*}
$$

This is equivalent to $\approx 2 \times 10^{-4}$ radians per century, or about 41 seconds of arc per century. A more accurate computation (also using general relativity, but with less severe approximations) predicts 43.0 seconds. The observed additional precession is $43.1 \pm 0.5$ seconds. Mercury's orbit actually precesses by around 575 Solar System, and a precession of 532 arcseconds per century could be deduced by including the effects of the other planets. The oblateness of the Sun gives an additional contribution of about 0.03 arcseconds per century. But until the advent of general relativity, the remaining 43 arcseconds per century could not be accounted for, however much people tried to tweak the existing (Newtonian) models. ${ }^{10}$

The reason for precession is partly an effect of the curvature of space. This causes the length of the orbiting trajectory to be shorter than our flat-space intuition tells us, so that we can think of there not being enough path length to complete a precession-free Newtonian orbit. However, the larger share of the effect is also due to the curvature effects on the time-coordinate part of the Schwarzschild geometry, so this really is a spacetime effect. ${ }^{11}$
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-101.jpg?height=340&width=363&top_left_y=640&top_left_x=1210)

Fig. 23.4 The geometry for the precession of the perihelion.
${ }^{10}$ Others had attempted to resolve the puzzle by tweaking Venus' mass up by $10 \%$, which was way more than was believable, or by positing a vast swarm lievable, or by positing a vast swarm
of asteroids close to Mercury, which had never been observed. As reported had never been observed. As reported
in Abraham Pais' biography, Einstein in Abraham Pais' biography, Einstein
was strongly affected by the remarkable agreement between the prediction of his theory and the astronomical observations. This instantly resolved a mystery which had stubbornly resisted solution since the 1850s. Einstein wrote to Paul Ehrenfest after his discovery, saying that "For a few days, I was beside myself with joyous excitement," and told another colleague that the discovery had given him heart palpitations. He also said that when he saw the agreement between his calculations and the unexplained observations, he had the feeling that something actually snapped in him.
${ }^{11}$ One lesson here is that although it is possible to think of curvature in terms of a rubber sheet model, this only tells us about spatial curvature and neglects half of the story. Indeed, the time part is the only part that matters for stationary particles. This point is examined in the exercises.

\section*{Chapter summary}
- Orbits of massive particles in the Schwarzschild geometry can be computed using the conservation of $\tilde{E}$ and $\tilde{L}$ and the identity $\boldsymbol{u}^{2}=$ -1 .
- The precession of the perihelion of Mercury is one of the classical tests of general relativity. The theory predicts results in excellent agreement with experiment.

\section*{Exercises}
(23.1) Consider a planet in a circular orbit, with radius $r$, around a spherically symmetric star in the $\theta=\pi / 2$ plane of a set of Schwarzschild coordinates.
(a) If the planet has a constant coordinate velocity $v=r \mathrm{~d} \phi / \mathrm{d} t$, use the line element to find the proper time taken to complete an orbit. Give your answer in terms of $v$ and $r$.
(b) Using the conservation laws in this chapter, show that the proper time to complete a circular orbit is given more generally as $\Delta \tau=$ $2 \pi r\left(\frac{r}{M}-3\right)^{\frac{1}{2}}$, and show that this is compatible with the answer in (a).
(23.2) Consider circular motion in a de Sitter geometry with metric
$$
\begin{equation*}
\mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left(\mathrm{~d} r^{2}+r^{2} \mathrm{~d} \phi^{2}\right) \tag{23.27}
\end{equation*}
$$

Calculate the proper time taken to complete an orbit, assuming constant velocity $v=a(t) R \mathrm{~d} \phi / \mathrm{d} t$ and a factor $a(t)=\mathrm{e}^{t}$.
(23.3) For a free particle at a constant radial coordinate in the Schwarzschild geometry, how does the $t$ variable vary with proper time $\tau$ ?
(23.4) A particle moves at a fixed radius in a Schwarzschild geometry at constant angular speed, such that it follows a world line given by
$$
\begin{array}{cc}
t=C \tau & r=r_{0} \\
\theta=\pi / 2 & \phi=\omega \tau,
\end{array}
$$
where $\tau$ is the proper time and $r_{0}, C$ and $\omega$ are constants.
(a) Use the constraint on the velocity to find an expression for $C$.
(b) Compute the components of the acceleration
of the particle.
(c) Under what circumstances does the acceleration vanish?
There are more complex (solved) problems of this sort in Blennow and Ohlsson, from which the preceeding ones have been adapted.
(23.5) The precession of the perihelion results from two curvature effects: $g_{t t}$ affects the rate of clock ticks and $g_{r r}$ leads to a curvature of space. This latter effect can be isolated.
(a) The spatial slice of the Schwarzschild spacetime with $t=$ const. and $\theta=\pi / 2$ has line element
$$
\begin{equation*}
\mathrm{d} s^{2}=\frac{\mathrm{d} r^{2}}{1-\frac{2 M}{r}}+r^{2} \mathrm{~d} \phi^{2} \tag{23.28}
\end{equation*}
$$

Show that this slice can be embedded in flat spacetime with cylindrical coordinates and
$$
\begin{equation*}
z(r)^{2}= \pm[8 M(r-2 M)] \tag{23.29}
\end{equation*}
$$

Hint: This is one of the cases we meet in Appendix $D$. The resulting surface is called Flamm's paraboloid [after Ludwig Flamm (1885-1964)]. The particle orbits, not in flat space, but instead on the surface represented by the paraboloid. This resembles, at least approximately, a cone that is tangent to the paraboloid as shown in Fig. 23.5. The distance from the origin in this approximation is $R$. For a circular orbit therefore, the apparent flat-space trajectory seems to have length $2 \pi R$, but this is too long when curvature is taken into account. This is because the real radius of the orbit is $2 \pi r=2 \pi R \cos \alpha$. We must therefore cut out a slice of angle $\delta$ to turn the circle of radius $R$ into the circle of radius $r$.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-103.jpg?height=212&width=463&top_left_y=407&top_left_x=476)

Fig. 23.5 (a) The flat-space radius $R$ and the real radius $R \cos \alpha$. (b) A wedge of angular size $\delta$ must be cut out of the large circle so that its circumference matches the smaller one.
(b) Using the fact that the tangent to the paraboloid is $\tan \alpha=\mathrm{d} z / \mathrm{d} r$, show that
$$
\begin{equation*}
\delta \approx \frac{2 \pi M}{r} \tag{23.30}
\end{equation*}
$$
(c) If the orbit is elliptical, argue that the perihelion must advance by an amount $\delta$ on every orbit. (d) How much of the perihelion shift is accounted for by the spatial curvature?
See the book by Moore for more details of this approach.
(23.6) Consider the flat-space (but curved spacetime) metric
$$
\begin{align*}
\mathrm{d} s^{2} & =-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\mathrm{d} r^{2} \\
& +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{23.31}
\end{align*}
$$

This can be used to calculate the contribution to the perihelion shift from the time component of the metric.
(a) Using the usual velocity identity, find an expression for $\mathrm{d} r / \mathrm{d} \phi$ for motion in the equatorial plane in this spacetime.
(b) Show that this leads to an equation of motion
$$
\begin{equation*}
\frac{\mathrm{d}^{2} u}{\mathrm{~d} \phi^{2}}+u=\frac{M \tilde{E}}{\tilde{L}^{2}(1-2 M u)^{2}} \tag{23.32}
\end{equation*}
$$
(c) Expand this in the limit $r \gg 2 M$ to show that for a circular orbit, with radius $r_{\mathrm{c}}$ we have
$$
\begin{equation*}
\left[1-4\left(\frac{M \tilde{E}}{\tilde{L}}\right)^{2}\right] u_{\mathrm{c}} \approx \frac{M \tilde{E}^{2}}{\tilde{L}^{2}} \tag{23.33}
\end{equation*}
$$
and use this to derive an approximate expression for $u_{\mathrm{c}}=1 / r_{\mathrm{c}}$ in terms of the constants in the problem.
(d) By expanding in terms of a perturbation to the circular orbit $u(\phi)=u_{c}+u_{c} w(\phi)$, find an equation for the perturbation $w(\phi)$.
(e) Hence, show that the perihelion shift for this metric accounts for $2 / 3$ of the total perihelion shift for the Schwarzschild metric.
Hint: You should obtain a harmonic oscillator equation for $w(\phi)$ with characteristic frequency $\omega_{0}$. Argue that the distance of closest approach is found when $\omega_{0} \phi=2 \pi n$, where $n$ is an integer, and use this to determine the shift $\Delta \phi$ between successive closest approaches. This method can be used to compute the perihelion shift for the Schwarzschild metric and is discussed in many books.
(23.7) Write a simple program to compute trajectories using eqn 23.15 for the case $\tilde{L} / M=4.3$.
Hint: Consider how to choose the sign in front of the equation, as this changes depending on whether the orbiting particle is heading towards or away from the point of closest approach.

\section*{24}

\section*{Photons in the Schwarzschild geometry}

\subsection*{24.1 Photon trajectories 254 \\ 24.2 Looking around 258 Chapter summary 260 \\ Exercises 261}
${ }^{1}$ In cylindrical polar coordinates, straight lines have the equation
$u=\frac{1}{r}=-\frac{m \cos \theta-\sin \theta}{c}$,
where $m$ is the gradient and $c$ the intercept on the $y$-axis.
${ }^{2}$ Johann Georg von Soldner (17761833). Other neglected predictions of gravitational effects on light rays were made by Henry Cavendish (1731-1810), John Michell (1724-1793) and Newton himself. Michell in particular has been described as one of the greatest unsung scientific heroes of all time, and was also the first person to suggest the existence of black holes, to explain the existence of double stars in terms of grav itation, ad to apply statistics to cos mology.
${ }^{3}$ Since in the weak-field limit we write $\sqrt{-g_{00}} \approx 1+2 \Phi$, we see that this coordinate speed $c^{\prime}$ varies with the gravitational potential $\Phi$.

Music is the arithmetic of sounds as optics is the geometry of light.
Claude Debussy (1862-1918)
In Newtonian gravitation, as we have described it, the photon does not interact with the gravitational field and so the paths of all photons in free space are straight lines. ${ }^{1}$ There had, however, been predictions of the bending of starlight around massive objects, notably by Johann Soldner ${ }^{2}$ who used Newton's corpuscular theory of light to predict that starlight passing close to the Sun would cause an apparent shift in the position of stars of around 0.8 arc seconds. In this chapter, we see what general relativity has to say about the gravitational interaction of light and matter.

\section*{Example 24.1}

So why should light rays be affected by gravity? To get an idea, we consider a static, diagonal metric. For the world line of a light ray, we have $0=c^{2} g_{00} \mathrm{~d} t^{2}+g_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j}$,

That is to say that the speed of light $c^{\prime}$ determined using the coordinate time varies. ${ }^{3}$ Although the world line of a light ray is a null geodesic in spacetime, it is not necessarily a geodesic in space. In fact, $\sqrt{-g_{00}}$ looks a lot like a refractive index and it can be shown that the spatial part of the trajectory obeys a version of Fermat's principle of least time that says
$$
\begin{equation*}
\delta \int\left(\frac{g_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j}}{-g_{00}}\right)^{\frac{1}{2}}=0 \tag{24.3}
\end{equation*}
$$

This provides at least some justification for the bending of light in a static gravitational field.

\subsection*{24.1 Photon trajectories}

In general relativity, the curvature of spacetime due to the presence of massive objects causes a photon trajectory to be deflected compared to its path in flat spacetime. One immediate complication with understanding the motion of photons is the impossibility of defining a proper
time $\tau$. Instead we must make use of an affine parameter $\lambda$ that marks off regular intervals along the world line of the photon. Key to computing the influence of gravitation on light is the observation that photons travel along null geodesics. These are paths whose velocity tangent vector $\boldsymbol{u}$ has the property
$$
\begin{equation*}
\boldsymbol{u} \cdot \boldsymbol{u}=g_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda} \frac{\mathrm{~d} x^{\nu}}{\mathrm{d} \lambda}=0 \quad \text { (photons) } \tag{24.4}
\end{equation*}
$$
where $\lambda$ is an affine parameter. We can write the null condition for photons in the Schwarzschild geometry in order to derive an equation of motion. Written out in terms of conserved quantities ${ }^{4} \tilde{E}$ and $\tilde{L}$, the null condition becomes
$$
\begin{equation*}
-\left(1-\frac{2 M}{r}\right)^{-1} \tilde{E}^{2}+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)^{2}+\frac{\tilde{L}^{2}}{r^{2}}=0 \tag{24.5}
\end{equation*}
$$

Multiplying by $(1-2 M / r)$, dividing by $\tilde{L}^{2}$ and following the algebra through, we end up with an effective energy equation for photons, just as we had for massive particles. We cast this in the form $\mathcal{E}=\mathcal{T}+W_{\text {eff }}$, where $\mathcal{T}$ is the kinetic-energy-like contribution.
The effective energy equation for photons is
$$
\begin{equation*}
\frac{1}{b^{2}}=\frac{1}{\tilde{L}^{2}}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)^{2}+W_{\mathrm{eff}}(r) \tag{24.6}
\end{equation*}
$$
where $b\left(\equiv \mathcal{E}^{-1 / 2}\right)=\tilde{L} / \tilde{E}$ and there is an effective potential $W_{\text {eff }}$ for photons of
$$
\begin{equation*}
W_{\mathrm{eff}}(r)=\frac{1}{r^{2}}\left(1-\frac{2 M}{r}\right) \tag{24.7}
\end{equation*}
$$

So the quantity $1 / b^{2}$ plays the role analogous to total effective energy $\mathcal{E}$, while the role of kinetic energy $\mathcal{T}$ is taken by $(\mathrm{d} r / \mathrm{d} \lambda)^{2} / \tilde{L}^{2}$.

\section*{Example 24.2}

In order to interpret the physical meaning of $b$, consider the geometry in Fig. 24.1. A photon is initially moving parallel to the $x$-axis, a perpendicular distance $d$ away from it, such that $y=d$ at infinity. This defines the impact parameter $d$. Far from any source of curvature the photon moves in a straight line, so for $r \gg 2 M$ we have
$$
\begin{equation*}
b=\left|\frac{\tilde{L}}{\tilde{E}}\right|=\frac{r^{2} \mathrm{~d} \phi / \mathrm{d} \lambda}{\mathrm{~d} t / \mathrm{d} \lambda}=r^{2} \frac{\mathrm{~d} \phi}{\mathrm{~d} t} \tag{24.8}
\end{equation*}
$$

At large $r$ we also have $\phi=\sin ^{-1}(d / r) \approx d / r$ and $\mathrm{d} r / \mathrm{d} t \approx-1$ so that
$$
\begin{equation*}
\frac{\mathrm{d} \phi}{\mathrm{~d} t}=\frac{\mathrm{d} \phi}{\mathrm{~d} r} \frac{\mathrm{~d} r}{\mathrm{~d} t}=\frac{d}{r^{2}} \tag{24.9}
\end{equation*}
$$

We conclude that $b=d$. That is, $b$ is the impact parameter for a light ray that reaches infinity.

The definition of $b$ makes sense physically in that the impact parameter is a measure of the angular momentum: it evaluates how far from the
${ }^{4}$ The conserved quantities followed from geometric considerations in the last chapter and so apply for photons as well as for massive particles, although we cannot parametrize the world lines of photons with the proper time.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-105.jpg?height=240&width=377&top_left_y=1092&top_left_x=1206)

Fig. 24.1 The geometry defining the impact parameter.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-106.jpg?height=392&width=363&top_left_y=608&top_left_x=315)

Fig. 24.2 The relativistic effective potential for photons.
${ }^{5}$ This step involves a little rearrangement of the indices
$$
g_{t t} \frac{\left(u^{t}\right)^{2}}{\left(u^{\phi}\right)^{2}}=\frac{g^{t t}\left(u_{t}\right)^{2}}{\left(g^{\phi \phi}\right)^{2}\left(u_{\phi}\right)^{2}}
$$
origin a trajectory is, but decreases with increasing energy, since there is less scattering of an energetic particle by the gravitational potential compared to one with less energy.
The most important quantity in understanding the trajectories of light rays is the effective potential $W_{\text {eff }}$ which is shown in Fig. 24.2. Unlike the analogous case for massive particles, the effective potential is independent of $\tilde{L}$ for photons. Like the relativistic $V_{\text {eff }}$, the potential $W_{\text {eff }}$ has a maximum; this occurs at $r_{0}=3 M$ where $W_{\text {eff }}\left(r_{0}\right)=\frac{1}{\left(27 M^{2}\right)}$. When an incoming photon has a large impact parameter, such that $1 / b^{2}<W\left(r_{0}\right)=1 / 27 M^{2}$, then the photon will always be deflected by the potential. However, if an incoming photon has a small enough impact parameter such that $1 / b^{2}>W\left(r_{0}\right)$ then the photon will spiral in towards $r=0$ where it is destroyed. Also unlike both the relativistic and Newtonian effective potentials for massive particles, $W_{\text {eff }}$ has no local minimum, so we should not expect any stable circular orbits for photons. However, at the maximum at $r_{0}$ circular orbits of light are possible, with impact parameter $b^{2}=27 M^{2}$, in a region of space called the photon sphere. Since this solution lies at the maximum of $W_{\text {eff }}$ these trajectories are unstable.
Using the same techniques as we employed for massive particles, we can come up with an equation of motion for the photons, again using the variable $u=M / r$.

\section*{Example 24.3}

Let's examine light rays in the $\theta=\pi / 2$ plane so we have that $u^{\theta}=0$. The equation providing the trajectory (the change in radius with $\phi$ ) is given by
$$
\begin{equation*}
\frac{\mathrm{d} r}{\mathrm{~d} \phi}=\frac{\mathrm{d} r}{\mathrm{~d} \lambda} \cdot \frac{\mathrm{~d} \lambda}{\mathrm{~d} \phi}=\frac{u^{r}}{u^{\phi}} \tag{24.10}
\end{equation*}
$$

Since all light rays are null, we have $\boldsymbol{u} \cdot \boldsymbol{u}=0$ and so
$$
\begin{equation*}
g_{t t}\left(u^{t}\right)^{2}+g_{r r}\left(u^{r}\right)^{2}+g_{\phi \phi}\left(u^{\phi}\right)^{2}=0 \tag{24.11}
\end{equation*}
$$
or, dividing through by $\left(u^{\phi}\right)^{2}$, we find
$$
\begin{equation*}
g_{t t} \frac{\left(u^{t}\right)^{2}}{\left(u^{\phi}\right)^{2}}+g_{r r}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \phi}\right)^{2}+g_{\phi \phi}=0 \tag{24.12}
\end{equation*}
$$

The first term can be rewritten ${ }^{5}$ using $\left(u_{t} / u_{\phi}\right)^{2}=1 / b^{2}$. On inserting the metric components, we have
$$
\begin{equation*}
\left(\frac{\mathrm{d} r}{\mathrm{~d} \phi}\right)^{2}=\left(1-\frac{2 M}{r}\right)\left[\frac{1}{b^{2}}\left(1-\frac{2 M}{r}\right)^{-1} r^{4}-r^{2}\right] . \tag{24.13}
\end{equation*}
$$

Now introduce the variable $u=M / r$ and find
or
$$
\begin{equation*}
\frac{M^{2}}{u^{4}} \cdot\left(\frac{\mathrm{~d} u}{\mathrm{~d} \phi}\right)^{2}=(1-2 u)\left[\frac{1}{b^{2}} \cdot(1-2 u)^{-1}\left(\frac{M}{u}\right)^{4}-\left(\frac{M}{u}\right)^{2}\right] \tag{24.14}
\end{equation*}
$$
$$
\begin{equation*}
\left(\frac{\mathrm{d} u}{\mathrm{~d} \phi}\right)^{2}=\frac{M^{2}}{b^{2}}-u^{2}+2 u^{3} \tag{24.15}
\end{equation*}
$$

Differentiating this latter equation, we find an equation of motion
$$
\begin{equation*}
\left(\frac{\mathrm{d}^{2} u}{\mathrm{~d} \phi^{2}}\right)+u=3 u^{2} \tag{24.16}
\end{equation*}
$$

The Newtonian analogue of this equation can be extracted from eqn 20.33 by setting $m=0$, yielding $u^{\prime \prime}+u=0$. The right-hand side of eqn 24.16 therefore provides the relativistic correction.

The equation of motion allows us to evaluate how light rays are deflected by gravitation. In fact, this effect is the second of our classical solarsystem tests of general relativity. The geometry for light deflection is shown in Fig. 24.3.

Example 24.4
Consider the equation of motion. In the limit $M / b \ll 1$, we expect the straight line solution
$$
\begin{equation*}
u_{0}=\frac{M}{b} \cdot \sin \phi, \tag{24.17}
\end{equation*}
$$
which solves the Newtonian equation $u_{0}^{\prime \prime}+u_{0}=0$. We can expand the equation of motion in terms of perturbations to this straight line as $u(\phi) \approx u_{0}(\phi)+u_{1}(\phi)$. Doing this in eqn 24.16, we find
$$
\begin{align*}
u_{1}^{\prime \prime}+u_{1} \approx 3 u_{0}^{2} & =3\left(\frac{M}{b}\right)^{2} \sin ^{2} \phi  \tag{24.18}\\
& =\frac{3}{2}\left(\frac{M}{b}\right)^{2}(1-\cos 2 \phi) \tag{24.19}
\end{align*}
$$
where we have retained only the largest term on the right (i.e. $u_{0}$ ) in the first of these expressions. The solution to this equation is given by
$$
\begin{equation*}
u_{1}=\frac{1}{2}\left(\frac{M}{b}\right)^{2}(3+\cos 2 \phi) \tag{24.20}
\end{equation*}
$$
so that we have a trajectory given by
$$
\begin{equation*}
u \approx\left(\frac{M}{b}\right) \sin \phi+\frac{1}{2}\left(\frac{M}{b}\right)^{2}(3+\cos 2 \phi) \tag{24.21}
\end{equation*}
$$

A deflection angle can be found by calculating the two angles for which $u=1 / r=0$. We find $u=0$ for
$$
\begin{equation*}
2 \sin \phi \approx-\left(\frac{M}{b}\right)(3+2 \cos \phi) \tag{24.22}
\end{equation*}
$$

Angles that approximately satisfy this condition are found at $\phi \approx$ $-2(M / b)$ and $\phi \approx \pi+2(M / b)$. Our conclusion is that the total deflection angle $\Delta \phi$ for light is therefore $4 M / b$.

\section*{Example 24.5}

For light that just grazes the Sun, ${ }^{6}$ we have $b=6.9 \times 10^{8} \mathrm{~m}$ and we compute a deflection (with constants restored) of
$$
\begin{equation*}
\Delta \phi=\frac{4 G M}{c^{2} b} \approx 8.6 \times 10^{-6} \mathrm{rad} \tag{24.23}
\end{equation*}
$$
which is equivalent to a prediction of 1.8 arc seconds (around twice Soldner's prediction based on Newtonian corpuscular theory). This was the prediction of general relativity that was tested by Arthur Eddington's 1919 observation of the shift in stellar positions during a solar eclipse, when starlight had passed close to the Sun would become temporarily visible. ${ }^{7}$ The expedition gave measurements that agreed with the prediction within estimated experimental uncertainties. This made Einstein an internationally celebrated figure. Some later analyses of the Eddington results have questioned the systematic errors involved in the measurements. Later high-precision measurements using radio sources have confirmed that the measured shifts are in excellent agreement with the predictions of general relativity (and the picture shown in Fig. 24.4 is a common way of illustrating the effect).
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-107.jpg?height=321&width=392&top_left_y=843&top_left_x=1201)

Fig. 24.3 The geometry for the deflection of a photon.
${ }^{6}$ The solar radius is usually taken to be $\approx 696,000 \mathrm{~km}$.
${ }^{7}$ For an account of the history, see J. Earman and C. Glymour, Historical Studies in the Physical Sciences 11, 175 (1980).
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-107.jpg?height=181&width=384&top_left_y=1586&top_left_x=1208)

Fig. 24.4 An illustration of how curved space bends starlight around the Sun.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-108.jpg?height=566&width=315&top_left_y=433&top_left_x=299)

Fig. 24.5 (a) Light bent by a gravitating mass is not focussed to a point. (b) An 'Einstein ring' is formed by the deflection of light originating from a source $s$ around a mass at $\ell$. The trajectory of the light is such that an observer at $O$ will trace the rays back via straight lines to form the circular image in the same plane as $s$.
${ }^{8}$ The consequences of this are examined in the exercises.

The deflection of light rays due to gravity allows us to treat gravitating masses a little like lenses. However, as discussed in the next example, the analogy is not an exact one.

Example 24.6
Consider the setup in Fig. 24.5(a) where light coming from infinity is bent by a gravitating mass at $\ell$ to reach the observer. Working in the small-angle approximation, light is detected a distance $D$ from $\ell$ where $D=b / \theta$. The deflection angle $\theta$ for a gravitating mass $M$ is $4 M / b$ and so we obtain
$$
\begin{equation*}
D=b^{2} / 4 M \tag{24.24}
\end{equation*}
$$

This shows that the light rays are focussed down to different points depending on how far they are from the axis. They are not, therefore, focussed down to a point. This is examined further in the exercises.

In the case that the source, lens, and observer are all in a line, the image takes the form of a ring, ${ }^{8}$ known as the Einstein ring, as demonstrated in Fig. 24.5(b).

\subsection*{24.2 Looking around}

The passage of light rays in a gravitational field determines what observers can see. To investigate this we must consider the momentum of photons. Recall that owing to photons being massless, the components of the photon momentum vector in flat spacetime are related to the velocity components of the photon via $p^{\mu}=E u^{\mu}$, where $E$ is the photon energy. Our first task is to update this for the curved spacetime we experience in the Schwarzschild geometry.

\section*{Example 24.7}

A momentum vector $\boldsymbol{p}$ for a particle of mass $m$ has components in the coordinate frame of
$$
\begin{equation*}
p^{\mu}=m \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} \tau}=m \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \cdot \frac{\mathrm{~d} t}{\mathrm{~d} \tau}=m \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \cdot u^{t} \tag{24.25}
\end{equation*}
$$

For motion in the Schwarzschild geometry, we have
$$
\begin{equation*}
\tilde{E}=\left(1-\frac{2 M}{r}\right) u^{t} \tag{24.26}
\end{equation*}
$$
and so we can write
$$
\begin{equation*}
p^{\mu}=\left(1-\frac{2 M}{r}\right)^{-1} m \tilde{E} \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \tag{24.27}
\end{equation*}
$$

The quantity $m \tilde{E}$ is the energy of the particle at infinity. If we write this quantity as $\varepsilon$ we have an expression
$$
\begin{equation*}
p^{\mu}=\left(1-\frac{2 M}{r}\right)^{-1} \varepsilon \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} t} \tag{24.28}
\end{equation*}
$$

This is well defined in the limit $m \rightarrow 0$ and so is suitable as a definition of the momentum components for the photon. The price we have paid is the inclusion of the coordinate $t$. In a coordinate system $(t, r, \theta, \phi)$, the momentum components become
$$
\begin{array}{cll}
p^{t}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1}, & p^{r}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} r}{\mathrm{~d} t}  \tag{24.29}\\
p^{\theta}=0, & p^{\phi}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} \phi}{\mathrm{~d} t} .
\end{array}
$$

We also have $b=r^{2}(1-2 M / r)^{-1} \mathrm{~d} \phi / \mathrm{d} t$ and, from the null condition for photons, we can write
$$
\begin{align*}
& 0=-\left(1-\frac{2 M}{r}\right)+\left(1-\frac{2 M}{r}\right)^{-1}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}+r^{2}\left(\frac{\mathrm{~d} \phi}{\mathrm{~d} t}\right)^{2}  \tag{24.30}\\
& \text { or }  \tag{24.31}\\
& \qquad 1=\left(1-\frac{2 M}{r}\right)^{-2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} t}\right)^{2}+\left(1-\frac{2 M}{r}\right) \frac{b^{2}}{r^{2}} \\
& \text { which gives us an expression for } \mathrm{d} r / \mathrm{d} t \text {. Collecting these facts, the final result for the }
\end{align*}
$$ momentum components of light is
$$
\begin{array}{cc}
p^{t}=\varepsilon\left(1-\frac{2 M}{r}\right)^{-1}, & p^{r}= \pm \varepsilon\left[1-\frac{b^{2}}{r^{2}}\left(1-\frac{2 M}{r}\right)\right]^{\frac{1}{2}},  \tag{24.32}\\
p^{\theta}=0, & p^{\phi}=\varepsilon \frac{b}{r^{2}} .
\end{array}
$$

The geometry for photons emitted from a point at some radius $r$ is shown in Fig 24.6. The most striking thing here is that photons emitted beyond a particular angle $\psi$ will have $\left(1 / b^{2}\right)>W\left(r_{0}\right)$ (see Fig. 24.7) and will therefore spiral into the origin, where they are destroyed. The function $\sin \psi$ equals the sideways component of the coordinate velocity of the photon. Referring to the figure, we see that, in the observer's local frame, this can be written as
$$
\begin{equation*}
\sin \psi=\frac{u^{\hat{\phi}}}{u^{\hat{t}}}=\frac{p^{\hat{\phi}}}{p^{\hat{t}}} \tag{24.33}
\end{equation*}
$$
where, in the second step, we've used $p^{\hat{\mu}}=E u^{\hat{\mu}}$ in the local frame. In order to use this, we need the conventional vielbein components for an orthonormal frame ${ }^{9}$ to shift components via
$$
\begin{equation*}
\sin \psi=\frac{\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}} p^{\phi}}{\left(\boldsymbol{e}_{t}\right)^{\hat{t}} p^{t}} \tag{24.34}
\end{equation*}
$$

Plugging in from the previous example we have
$$
\begin{equation*}
\sin \psi=\frac{b}{r}\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{24.35}
\end{equation*}
$$

The geometry is such that if the angle $\psi$ is greater than a critical angle $\psi_{\mathrm{c}}$ then $1 / b^{2}>W\left(r_{0}\right)$ and the photon will spiral into the origin. This critical angle occurs when $1 / b^{2}=W\left(r_{0}\right)$, or $b=\sqrt{27} M$. [When the equality holds we expect the photons to (unstably) orbit the central mass.] Substituting, we conclude that those photons with angles greater than $\sin \psi_{\mathrm{c}}=(\sqrt{27} M / r)(1-2 M / r)^{\frac{1}{2}}$ are condemned to fall into the central mass, where they are destroyed. Photons with angles $\psi<\psi_{\text {c }}$
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-109.jpg?height=543&width=357&top_left_y=433&top_left_x=1219)

Fig. 24.6 The geometry for photons emitted from a point at radius $r$.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-109.jpg?height=388&width=377&top_left_y=1095&top_left_x=1206)

Fig. 24.7 A photon with $1 / b^{2}>$ $W_{\text {eff }}\left(r_{0}\right)$ will spiral into the origin of the coordinate system.
${ }^{9}$ The vielbein components for $\theta=\pi / 2$ are
$\left(\boldsymbol{e}_{t}\right)^{\hat{t}}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}}$,
$\left(\boldsymbol{e}_{r}\right)^{\hat{r}}=\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}}$,
$\left(e_{\theta}\right)^{\hat{\theta}}=r$,
$\left(\boldsymbol{e}_{\phi}\right)^{\boldsymbol{\phi}}=r$.
escape, but as the radius $r$ decreases, more and more of the photons are captured since the critical angle becomes smaller. The critical angle vanishes for $r=2 M$ and all photons are captured.

The Schwarzschild geometry is invariant with respect to time reversal. That is, if we play a film of the trajectories in reverse, the geometry is unchanged and so this same physics largely applies to the photons received by an observer at position $r$. The exception is the behaviour of those photons that were destroyed by spiralling into the origin: when we play the film backwards, these photons are not emitted and that part of space at angles $\psi>\psi_{c}$ is black. We will see the consequence of this in the next chapter, whose subject is black holes.

\section*{Chapter summary}
- Photons are deflected by gravitating objects in general relativity.
- An effective-energy equation allows deflection to be calculated.
- The deflection of light by gravitating stars is one of the classical tests of general relativity. The theory predicts results in excellent agreement with experiment.

It is useful at this stage to summarize the equations for motion in the Schwarzschild geometry for massive particles and for photons, where we assume the motion takes place in the equatorial plane.
\begin{tabular}{ll}
\hline & Massive particles \\
\hline Useful constants & $\tilde{E}=\left(1-\frac{2 M}{r}\right) \frac{\mathrm{d} t}{\mathrm{~d} \tau}$ \\
& $\tilde{L}=r^{2} \frac{\mathrm{~d} \phi}{\mathrm{~d} \tau}$ \\
\hline$\frac{\mathrm{~d} r}{\mathrm{~d} \tau}$ & $\pm\left[\tilde{E}^{2}-\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}}{r^{2}}\right)\right]^{\frac{1}{2}}$ \\
\hline$\frac{\mathrm{~d} r}{\mathrm{~d} t}$ & $\pm\left(1-\frac{2 M}{r}\right)\left[1-\frac{1}{E^{2}}\left(1-\frac{2 M}{r}\right)\left(1+\frac{\tilde{L}^{2}}{r^{2}}\right)\right]^{\frac{1}{2}}$ \\
\hline$\frac{\mathrm{~d} \phi}{\mathrm{~d} t}$ & $\frac{\tilde{L}}{r^{2} E}\left(1-\frac{2 M}{r}\right)$ \\
\hline & Photons \\
\hline$\frac{\mathrm{Useful} \text { constant }}{}$ & $b=\frac{\tilde{L}}{E}=r^{2}\left(1-\frac{2 M}{r}\right)^{-1} \frac{\mathrm{~d} \phi}{\mathrm{~d} t}$ \\
\hline$\frac{\mathrm{~d} r}{\mathrm{~d} t}$ & $\pm\left(1-\frac{2 M}{r}\right)\left[1-\left(1-\frac{2 M}{r}\right) \frac{b}{}^{2} r^{\frac{1}{2}}\right]^{\frac{1}{2}}$ \\
\hline$\frac{\mathrm{~d} \phi}{\mathrm{~d} t}$ & $\frac{b}{r^{2}}\left(1-\frac{2 M}{r}\right)$ \\
\hline
\end{tabular}

\section*{Exercises}
(24.1) Confirm eqns 24.6 and 24.7 .
(24.2) Confirm that eqn 24.21 solves eqn 24.19.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-111.jpg?height=412&width=549&top_left_y=681&top_left_x=427)

Fig. 24.8 Ray diagram for Exercise 24.3, showing a source $s$, a lensing mass $\ell$ and an observer $O$.
(24.3) Consider the optical diagram in Fig. 24.8.
(a) Using the small-angle approximation, show that
$$
\begin{equation*}
\theta_{\mathrm{i}}=\theta_{\mathrm{s}}+\frac{\theta_{\mathrm{E}}^{2}}{\theta_{\mathrm{i}}} \tag{24.36}
\end{equation*}
$$
where we have defined the Einstein angle $\theta_{\mathrm{E}}^{2}=$ $4 M D_{\ell \mathrm{s}} / D_{\ell} D_{\mathrm{s}}$.
(b) Find an expression of the angular size of the Einstein ring for the case that the source is at infinity and the source, lensing mass and observer are collinear.
(24.4) A light ray travels along a radial line in a spacetime described by the Robertson-Walker line element
$$
\mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2}\left(\frac{\mathrm{~d} r^{2}}{1-k r^{2}}+r^{2} \mathrm{~d} \Omega^{2}\right)
$$
where $\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}$.
(a) Show that, when considering trajectories along the ray, the metric can be rewritten as
$$
\begin{equation*}
\mathrm{d} s^{2}=-\mathrm{d} t^{2}+a(t)^{2} \mathrm{~d} \chi^{2} \tag{24.38}
\end{equation*}
$$
where $\chi$ is a function of $r$ only.
(b) Show that the velocity component $u^{\chi}$ for this metric can be written as
$$
\begin{equation*}
u^{\chi}=\frac{C}{a(t)^{2}}, \tag{24.39}
\end{equation*}
$$
where $C$ is a constant.
(c) Show further that the affine parameter evolves along the light ray according to
$$
\begin{equation*}
\lambda=\int \frac{1}{C} \frac{a(t)^{2} \mathrm{~d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}} \tag{24.40}
\end{equation*}
$$

The knowledge of how the affine parameter varies along a geodesic is often useful. See Exercise 26.4 for an example.

\section*{25}
25.1 The surface $r=2 M \quad 264$ 25.2 The tortoise coordinate 265 25.3 Death of an astronaut 266 25.4 Looking around near a black
hole hole
25.5 Gravitational collapse 268 Chapter summary 270 Exercises 270
${ }^{1}$ John Michell, and subsequently Pierre-Simon Laplace (1749-1827), had suggested the possibility of a dark star whose gravitation would be such that not even light would have the necessary velocity to escape the star's surface. The modern concept of a black hole was investigated by Robert Oppenheimer (1904-1967) and coworkers in the late 1930s, who proposed that the gravitational collapse of neutron stars could potentially cause them to become black holes (although the term 'black hole' was not used at this point; it was popularized by John Wheeler in the 1960 s). The key paper Wheeler in the 1960 s ). The key paper
on gravitational collapse, co-authored on gravitational collapse, co-authored
by Oppenheimer and Hartland Snyder (1913-1962) was published in 1939 on the eve of World War II, a conflict in which Oppenheimer would, of course, play a significant role.
${ }^{2}$ The radius $r_{\mathrm{S}}$ corresponds to a spherical shell of singularities in (3+1)dimensional spacetime. Since we usually draw space on a two-dimensional page, it is sometimes called a a 'ringlike' singularity. The Schwarzschild singularity is sometimes also called the 'Hadamard catastrophe', as this is what Einstein jokingly called it, referring to the Jacques Hadamard's (1865-1963) the Jacques Hadamard's (1865-1963)
suggestion that it could be regarded resuggestion
alistically.
${ }^{3}$ That is, frequency goes to zero since an oscillation takes infinite time.

\section*{Black holes}

When black clouds envelop stars which shone bright,
They can no longer pour forth their light.
Boethius (c.480-c.524/6) The Consolation of Philosophy
A black hole ${ }^{1}$ is a part of spacetime where gravitational effects are so severe that light cannot escape from the inside of the region. It is likely that a black hole is the state of matter that, if massive enough, a star can adopt at the end of its life. Spacetime with a non-rotating black hole at its centre is spherically symmetric and must therefore, by Birkhoff's theorem, be described by the Schwarzschild metric.

The Schwarzschild geometry for the spacetime outside a gravitating object with mass $M$ is represented by the line element
$$
\begin{equation*}
\mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{25.1}
\end{equation*}
$$

It is notable that the Schwarzschild line element looks to be very badly behaved at two coordinate values: $r=0$ and also at $r=r_{\mathrm{S}}=2 M$, where $r_{\mathrm{S}}$ is known as the Schwarzschild radius. ${ }^{2}$ On the three-dimensional spherical surface defined by $r_{\mathrm{S}}$, trouble comes from two places: (i) the vanishing of the first term tells us that clocks at the Schwarzschild radius measure no time, which causes light signals to be infinitely redshifted; ${ }^{3}$ (ii) the divergence of the second term means that $\partial s / \partial r$ is infinite: a small change in $r$ leads to a divergent change in the invariant interval $\mathrm{d} s$. When Schwarzschild's solution was first discussed, it was expected that since the singularity at $r=0$ was hidden deep inside objects like stars, its effects should be unobservable. The singularity at $r_{\mathrm{S}}$ originally had a similar status: if we restore units and plug numbers in, we find
$$
\begin{equation*}
r_{\mathrm{S}}=\frac{2 G M}{c^{2}} \approx \frac{3 M}{M_{\odot}} \mathrm{km} \tag{25.2}
\end{equation*}
$$
where $M_{\odot} \approx 2 \times 10^{30} \mathrm{~kg}$ is the solar mass. For objects with masses of order a solar mass, it appeared that this badly behaved point was buried well within the innards of the star. However, the discovery of incredibly massive, dense objects with very strong gravitational fields has forced relativists to confront the details of the singular coordinate $r_{\mathrm{S}}$ head on. Specifically, the mass distribution for a black hole is actually confined to an infinitely dense point at the origin, with the result that the region $0<r<2 M$ lies outside of the mass that forms the hole.

It is the behaviour of massive particles and of light close to the radius $r_{\mathrm{S}}=2 M$ that leads to many of the most notable features of a black hole. We shall see that although the metric at $r_{\mathrm{S}}=2 M$ has several interesting features, it does not lead to any bad behaviour in the local physical fields. The curvature of spacetime, for example, varies smoothly over this region. An astronaut travelling towards the origin would not notice anything strange in their trajectory as they passed this point, with their watch continuing to tick regularly as they cross the surface $r=r_{\mathrm{S}}$ in a finite interval of proper time. (Indeed, we saw in Chapter 22 that any point in the geometry can be reached, via a radial plunge, in a finite interval of proper time.) In fact, the threat of an infinity at $r_{\mathrm{S}}$ is actually the consequence of our choice of coordinates in the Schwarzschild geometry and much of our task of the next chapter will be to identify a set of coordinates that elucidates the behaviour of light and particles near black holes. In this chapter, our task is to examine the behaviour of geodesics that pass close to $r=2 M$ using the Schwarzschild geometry. ${ }^{4}$

\section*{Example 25.1}

What is the evidence for the existence of black holes? As they cannot be directly seen (since, of course, light cannot escape their vicinity), the evidence for them is necessarily somewhat indirect. The observational evidence falls into four main categories.
- Stars close to the centre of the Milky Way have proper motions that indicate that they are whirling around a very compact radio source, believed to be a supermassive black hole of mass around $4 \times 10^{6} M_{\odot}$, named Sagittarius $A^{*}$. The accretion disc around this was imaged by the Event Horizon Telescope in 2022 (see Fig. 25.1).
- Binary star systems that are strong sources of $X$-rays. The mass of an unseen partner in a binary star system can be determined by its influence on the visible partner and can be inferred to be too massive to be a white dwarf or a neutron star. The X-ray emission occurs owing to the mass at the centre of the unseen part of the binary accreting an energetic disc of massive material from its partner. This matter orbits around the mass (at a distance larger than the event horizon, see below). Owing to the large velocities of the matter in the accretion disc, (which are caused by the large amount of angular momentum involved in the formation of the disc), the material in the disc has a temperature high enough to radiate strongly in the X-ray region of the electromagnetic spectrum.

The most famous example is Cygnus-X1, an X-ray source found in the Cygnus constellation. The object is partnered with the blue supergiant HDE 226868 in a binary system. The X-ray emission is consistent with a source with diameter smaller than several hundred kilometres. The mass at the centre of Cygnus-X1 is estimated to be $\approx 15 M_{\odot}$, which is far too large to be a neutron star.
- Anomalous behaviour close to the centre of a galaxy, that can be explained by an enormous, massive, but unseen, object. Such objects can also be sources of large amounts of energy. Quasars ${ }^{5}$ are very luminous (and therefore hugely energetic) objects that lie at the centre of galaxies. They are thought to comprise a supermassive black hole (i.e. black holes of millions to billions of $M_{\odot}$ ) surrounded by an accretion disc. The most distant quasar known is ULAS J1342+0928, which is expected to comprise the oldest known supermassive black hole, with a mass estimated at $8 \times$ $10^{8} M_{\odot}$.
- Gravitational wave observations that can be explained by the collision of black holes. We will return to these when we discuss gravitational waves in Chapter 46.
${ }^{4}$ It's perhaps worth stressing that the picture of a black hole as an entity relentlessly sucking in all of the matter in the Universe is not an accurate one. Its exterior geometry is no different to that of any other star. As long as an observer has access to some means of propulsion, they can always escape a black hole (although, as we shall see, this is as long as they never get closer than $r=r_{\mathrm{S}}$ ).
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-113.jpg?height=392&width=392&top_left_y=842&top_left_x=1201)

Fig. 25.1 An image of the supermassive black hole, Sagittarius A*, at the centre of the Milky Way, captured by an array of eight radio observatories distributed around Earth, networked together as the Event Horizon Telescope (EHT). The image shows a dark central region (called a shadow) surrounded by a bright ring-like structure. A mass of $4 \times 10^{6} M_{\odot}$ is shoehorned into a region of space of diameter of only 0.3 AU. (Courtesy ESO.)
${ }^{5}$ This curious word is a contraction of quasi-stellar radio sources. As the name suggests, they were originally thought to be stars. In spite of this, their large energies, large redshifts and the jets that they can emit are good evidence that they cannot be stars.
${ }^{6}$ As we saw in Chapter 22, coordinate time and the proper time of the observer travelling towards the surface are very different.
$t$
$i$
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-114.jpg?height=144&width=369&top_left_y=1294&top_left_x=317)

Fig. 25.2 Starting at large $r$, light cones eclipse as $r$ decreases on the approach to a black hole, before turning over for $r<r_{\mathrm{S}}$.
${ }^{7}$ This is a vivid manifestation of the lack of any metric significance of coordinates in general relativity: a variable called $t$ has no more claim to tell the time than one called $u$ or $\chi$.
${ }^{8}$ This section should therefore really have been called 'Inside and after the surface $r=2 M$,

\subsection*{25.1 The surface $r=2 M$}

Let's consider the regions close to the spherical surface defined by $r_{\mathrm{S}}=$ $2 M$. First, we look at the geometry outside this surface, where $r>2 M$. Light cones in the Schwarzschild geometry, like all light cones, have $\mathrm{d} s^{2}=0$, so we have from eqn 25.1 that, for constant $\theta$ and $\phi$, they are described by
$$
\begin{equation*}
\frac{\mathrm{d} t}{\mathrm{~d} r}= \pm\left(1-\frac{2 M}{r}\right)^{-1} \tag{25.3}
\end{equation*}
$$

The light cones have slope $\pm 1$ for large $r$ where spacetime becomes (asymptotically) flat. As we approach the surface $r_{\mathrm{S}}$ from $r>2 M$ (denoted $r \rightarrow 2 M^{+}$below) the slope of the light cone approaches $\pm \infty$ (meaning that the null surfaces of the cone point vertically in a standard $r$ vs. $t$ plot), closing up (or eclipsing) as shown in Fig. 25.2. As always, massive particles cannot travel faster than light, so are confined to the timelike part of the cone. The consequence of the narrowing of the light cones as $r \rightarrow 2 M^{+}$is that the trajectories of particles become more vertical in the $r$ - $t$ plane, which is to say that $r$ cannot change as much for a given time interval. Therefore, as the particle approaches $r_{\mathrm{S}}=2 M$ it seems to take longer and longer in coordinate time $t$ to make a change in $r$. In fact, this is the effect we saw in Chapter 22 where it takes infinite amount of coordinate time to reach the surface at $r_{\mathrm{S}}$ during a radial plunge. ${ }^{6}$

Let's now examine the geometry of spacetime just inside the surface at $r_{\mathrm{S}}$. Introducing the coordinate $\varepsilon$ via $r=2 M-\varepsilon$, the line element can be written as
$$
\begin{equation*}
\mathrm{d} s^{2}=\frac{\varepsilon}{2 M-\varepsilon} \mathrm{d} t^{2}-\frac{2 M-\varepsilon}{\varepsilon} \mathrm{d} \varepsilon^{2}+(2 M-\varepsilon)^{2} \mathrm{~d} \Omega^{2} \tag{25.4}
\end{equation*}
$$

We see that if we fix $t, \theta$ and $\phi$ at constant values, the interval is $\mathrm{d} s^{2}=$ $-(2 M-\varepsilon) \mathrm{d} \varepsilon^{2} / \varepsilon$. Remarkably, despite our having fixed things so that $\mathrm{d} t=0$, this interval has $\mathrm{d} s^{2}<0$, which is to say that it is timelike! This implies that $\varepsilon$ (the 'radial' coordinate inside the surface) is timelike, rather than spacelike as we expect for spatial coordinates. A timelike $\varepsilon$ is doomed to constantly increase (like time in flat spacetime), causing the radial coordinate $r$ to decrease until eventually we meet the singularity at $r=0$. By the same token, the 'time' coordinate $t$ in eqn 25.4 has become spacelike ${ }^{7}$ inside the surface $r_{\mathrm{S}}=2 M$.

A physical consequence of this topsy-turvy state of affairs can be seen by considering an astronaut at radius $r<2 M$ who sends out a photon. As shown in Fig. 25.2, the light cones inside $r=2 M$ tip over, spitting particles towards the singularity at $r=0$. That is to say that the photon must go forward in time, which means $r$ decreases and the photon has no option but to fall towards the origin. This means that for any observer within the event horizon, the future points towards $r=0$ or, to sloganize: the future is inwards. ${ }^{8}$ Photons inside the surface $r=2 M$ are therefore trapped and, as a consequence, so are all massive particles. With the impossibility of photons escaping the $r_{\mathrm{S}}$ surface, it is impossible
for anything inside $r_{\mathrm{S}}$ to be seen. In the same way that we cannot see beyond the Earth's horizon, we call the surface at $r_{\mathrm{S}}$ the event horizon of the black hole.
Taking the above considerations into account, we can plot the light cones in Schwarzschild coordinates, which are shown in Fig. 25.3. The light cones have the alarming feature that null trajectories appear singular at $r_{\mathrm{S}}$ with incoming trajectories shooting off to infinite $t$ before returning. It will turn out that this is not, in fact, real physical behaviour and in the next chapter we shall use an alternative coordinate description to understand the illusory nature of this feature.

\subsection*{25.2 The tortoise coordinate}

The investigation of black holes in general relativity is, in large part, an exercise in finding the right coordinates with which to describe them. In evaluating the coordinate time $t$ measured by some observer, we will often have cause to integrate equations like eqn 25.3 . Integrals of the form
$$
\begin{equation*}
\Delta t=\int \frac{\mathrm{d} r}{1-\frac{2 M}{r}} \tag{25.5}
\end{equation*}
$$
lead to logarithmic contributions to the coordinate-time interval. To understand these logarithms, an interesting coordinate that we can employ is the tortoise coordinate $r^{*}$, defined by
$$
\begin{equation*}
\frac{\mathrm{d} r^{*}}{\mathrm{~d} r}=\left(1-\frac{2 M}{r}\right)^{-1} \tag{25.6}
\end{equation*}
$$
so that we find (Fig. 25.4)
$$
\begin{equation*}
r^{*}=r+2 M \ln \left|\frac{r}{2 M}-1\right|+\text { const. } \tag{25.7}
\end{equation*}
$$

\section*{Example 25.2}

This coordinate is named in honour of the story of Achilles ${ }^{9}$ who races the tortoise and, despite moving at a high speed, apparently can never overtake it. This is because when Achilles catches up with the tortoise, the tortoise has moved to a new location. Achilles catches up again, but again, by the time he does, the tortoise has moved a bit further along the race track.
In our coordinates, we imagine a rather contrived race between Achilles, whose separation from the tortoise is given by $d=r-r_{\mathrm{S}}$, so that when $r=r_{\mathrm{S}}=2 M$, Achilles expects to pass the tortoise. We track progress in the race using the coordinate $r^{*}$ which one can think of as a curious sort of race timer, that starts at positive values of $r^{*}$ and monotonically decreases, with the possibility of taking all values $-\infty \leq r^{*} \leq \infty^{10}$ We start with $r^{*}$ taking a large, positive value, corresponding to the athletes being well separated (i.e. large $d=r-r_{\mathrm{S}}$ ). As we time-evolve the race, we allow the timer $r^{*}$ to decrease, and Achilles closes in on the tortoise, with $r$ approaching $r_{\mathrm{S}}$. As the race coordinate $r^{*}$ becomes large and negative and $r$ gets close to $r_{\mathrm{S}}$, we see from eqn 25.6 that $r$ changes more and more slowly with the race time $r^{*}$, since the race velocity $\mathrm{d} r / \mathrm{d} r^{*} \rightarrow 0$. We continue to time-evolve the race, making the $r^{*}$ coordinate more and more negative. However, Achilles never reaches the tortoise.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-115.jpg?height=340&width=384&top_left_y=412&top_left_x=1211)

Fig. 25.3 Light cones in the Schwarzschild geometry.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-115.jpg?height=366&width=377&top_left_y=969&top_left_x=1206)

Fig. 25.4 A plot of eqn 25.7.
${ }^{9}$ Achilles was a hero of the Trojan war and has a starring role in Homer's Iliad. Achilles is played by Brad Pitt in the film Troy, and even appears in the title of the Led Zeppelin track 'Achilles Last Stand'. The philosopher Zeno of Elea (Fifth century bc) picked Achilles for his paradox because, with regards to land speed, the slowly moving tortoise would be expected to be no match for the supremely athletic Achilles.
${ }^{10}$ Since coordinates have no intrinsic metric significance, we don't worry about exactly what this means in terms of the workings of the clock.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-116.jpg?height=409&width=358&top_left_y=389&top_left_x=317)

Fig. 25.5 Motion of an astronaut falling into a black hole as a function of proper time.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-116.jpg?height=403&width=377&top_left_y=905&top_left_x=302)

Fig. 25.6 Motion of an astronaut falling into a black hole as a function of Schwarzschild coordinate time.
${ }^{11}$ The equation of motion for the separation of two point is given, in the orthonormal frame, by
$$
\frac{\mathrm{d}^{2} \chi^{\hat{i}}}{\mathrm{~d} \tau^{2}}=-R_{\hat{\tau} \hat{j} \hat{\tau}}^{\hat{i}} \chi^{\hat{j}}
$$

The components of the Riemann tensor are
$$
\begin{array}{r}
R_{\hat{\gamma} \hat{r} \hat{r} \hat{r}}=-\frac{2 M}{r^{3}}, \\
R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}}=+\frac{2 M}{r^{3}}, \\
R_{\hat{\gamma} \hat{\theta} \hat{\gamma} \hat{\theta}}=R_{\hat{\tau} \hat{\phi} \hat{\gamma} \hat{\phi}}=+\frac{M}{r^{3}}, \\
R_{\hat{r} \hat{\theta} \hat{r} \hat{\theta}}=R_{\hat{r} \hat{\phi} \hat{\phi} \hat{\phi}}=-\frac{M}{r^{3}} .
\end{array}
$$

Notice that the curvature has no notable behaviour at $r=2 M$.

The tortoise coordinate is useful as it can be used to prevent the eclipse of the light cones that we saw occurring in the Schwarzschild coordinates as $r \rightarrow 2 M^{+}$owing to the property that $\mathrm{d} t / \mathrm{d} r= \pm(1-2 M / r)^{-1} \rightarrow \infty$. The solution is to note that the light cones can be described by setting
$$
\begin{equation*}
t= \pm r^{*}+\text { const., } \tag{25.8}
\end{equation*}
$$
where $r^{*}$ is the tortoise coordinate. If we then rewrite the metric we obtain
$$
\begin{equation*}
\mathrm{d} s^{2}=\left(1-\frac{2 M}{r}\right)\left[-\mathrm{d} t^{2}+\left(\mathrm{d} r^{*}\right)^{2}\right]+r^{2} \mathrm{~d} \Omega^{2} \tag{25.9}
\end{equation*}
$$
with $r$ a function of $r^{*}$. This solves the problem of the light cones closing at $r=2 M$ and it also prevents the metric having any bad behaviour around $r=2 M$. Nonetheless, it is not an easy set of coordinates to work with, since the point $r=2 M$ now occurs at infinite $r^{*}$. We return to the problem of finding better coordinates in the next chapter.

\subsection*{25.3 Death of an astronaut}

We observe an astronaut undergoing a radial plunge towards a black hole. She has a torch that emits light pulses. We saw in the last chapter that, from her point of view, the proper time during a radial plunge changes with the $r$ coordinate via
$$
\begin{equation*}
\tau=\tau_{*}+\frac{2}{3} \frac{r^{\frac{3}{2}}}{r_{\mathrm{S}}^{\frac{1}{2}}} \tag{25.10}
\end{equation*}
$$

The proper time for the astronaut to fall through the horizon is finite, and is shown by a solid line in Fig. 25.5. The astronaut notices nothing special about her clock as she passes the point of no return at $r_{\mathrm{S}}$.
However, from our point of view as observers at spatial infinity, our spacetime is flat and our proper time coincides with the coordinate time $t$. Using the result from the last chapter, the coordinate time during a radial plunge evolves according to
$$
\begin{equation*}
t=t_{*}+r_{\mathrm{S}}\left[-\frac{2}{3}\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{3}{2}}-2\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{1}{2}}+\ln \left|\frac{\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{1}{2}}+1}{\left(\frac{r}{r_{\mathrm{S}}}\right)^{\frac{1}{2}}-1}\right|\right] \tag{25.11}
\end{equation*}
$$

This quantity diverges as $r \rightarrow r_{\mathrm{S}}$ and so, we see the astronaut taking an infinite amount of time to cross the boundary. This is shown in Fig. 25.6.

Example 25.3
What does the astronaut feel as she falls? ${ }^{11}$ We evaluated the forces that act on an astronaut in the Schwarzschild geometry in the orthonormal frame in Chapter 11. These are

The astronaut is stretched out like spaghetti in the $\hat{r}$ direction, and compressed in the $\hat{\theta}$ and $\hat{\phi}$ directions. Notice that there are no special forces that the astronaut feels as she passes $r_{\mathrm{S}}$. Nevertheless, the astronaut is doomed to be stretched out (and compressed) to death by the forces whose magnitudes all diverge as $r \rightarrow 0$.

The last example suggests that, whatever the status of the bad behaviour at $r_{\mathrm{S}}$, we should take the singular behaviour of the metric at $r=0$ very seriously. This is indeed the case as the point $r=0$ does represent a physical singularity. Any body that meets the point $r=0$ must be destroyed by the infinite forces at this point. Meeting this point is the fate of anything that finds itself within the event horizon of a black hole.

What about the light pulses that the astronaut emits? Light rays travel along null geodesics and so, assuming the pulses are emitted radially, the interval between two events on the world line of the photons emitted by the astronaut are given by
$$
\begin{equation*}
0=-\left(1-\frac{r_{\mathrm{S}}}{r}\right) \mathrm{d} t^{2}+\frac{\mathrm{d} r^{2}}{1-\frac{r_{\mathrm{S}}}{r}} \tag{25.13}
\end{equation*}
$$
or
$$
\begin{equation*}
\frac{\mathrm{d} t}{\mathrm{~d} r}= \pm \frac{1}{1-\frac{r \mathrm{~s}}{r}} \tag{25.14}
\end{equation*}
$$

The photons of interest are those travelling radially outwards, so we choose the $+\operatorname{sign}$ (i.e. $r$ increases as $t$ increases). The journey time for a photon emitted at $\left(t_{1}, r_{1}\right)$ and detected at $\left(t_{2}, r_{2}\right)$ is
$$
\begin{align*}
t_{2}-t_{1} & =\int_{r_{1}}^{r_{2}} \frac{\mathrm{~d} r}{1-\frac{r_{\mathrm{S}}}{r}} \\
& =\left(r_{2}-r_{1}\right)+r_{\mathrm{S}} \ln \left(\frac{r_{2}-r_{\mathrm{S}}}{r_{1}-r_{\mathrm{S}}}\right) \tag{25.15}
\end{align*}
$$

The coordinate-time interval is therefore corrected from the usual (flat space) value ( $r_{2}-r_{1}$ ), with the journey time increased by the logarithmic correction (Fig. 25.7). The logarithm gets larger and larger as the point of emission $r_{1}$ gets closer to $r_{\mathrm{S}}$, causing the interval to diverge. As seen by us distant observers, the light pulses from the astronaut become less and less frequent. If we are able to see the astronaut during her descent then she will appear to never quite reach the horizon owing to the coordinate time interval getting larger and larger as we saw above. In addition to the light pulses emitted from the astronaut becoming less frequent, they will also become dimmer as light is severely redshifted by the gravitational effect. ${ }^{12}$
${ }^{12}$ Since this observer is falling, we cannot simply use the result from eqn 22.16 which assumes a stationary observer. In fact, we will examine just how severe the redshift in the next chapter after we have identified a more suitable set of coordinates to describe the situation.

\subsection*{25.4 Looking around near a black hole}

What does the astronaut see as she plunges towards the hole? Recall from the previous chapter that there is a critical angle for photons
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-117.jpg?height=431&width=375&top_left_y=412&top_left_x=1204)

Fig. 25.7 Equation 25.15, giving an interval $\Delta t=t_{2}-t_{1}$, plotted as a function of $r_{2}$ for given values of $r_{1}$.
${ }^{13}$ Interestingly, since when the equality holds we expect the photons to or bit the central mass, photons emitted from an observer at $r=3 M$ and $\psi_{c}$ can be observed by that same observer after completing an orbit. This implies that the astronaut can see the back of her head.
${ }^{14}$ The use of the vielbein components here is equivalent to the usual rule that $E=-\boldsymbol{p} \cdot \boldsymbol{u}_{\text {obs }}$ that we used to work out the gravitational redshift in Chapter 22. Recall also that our conventional vielbein corresponds to a stationary observer's local rest frame.
${ }^{15}$ That is, $p_{t}(r)=(1-2 M / r) p^{t}(r)$ is conserved. Therefore, $p^{t}(\infty)=\hbar \omega_{\infty}=$ $(1-2 M / r) p^{t}(r)$, so $p^{t}(r)=\hbar \omega_{\infty}(1-$ $2 M / r)^{-1}$.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-118.jpg?height=404&width=372&top_left_y=1047&top_left_x=319)

Fig. 25.8 Gravitational collapse with two spatial dimensions suppressed. Each point represents a 2-sphere.
${ }^{16}$ Subrahmanyan Chandrasekhar (1910-1995) computed that a white dwarf with a mass $\gtrsim 1.4 M_{\odot}$ (the Chandrasekhar limit) would be unstable to further gravitational collapse. This results in a neutron star or, if the star is still more massive, a black hole. The most massive known neutron star is $\approx 2.4 M_{\odot}$; the least massive black hole is thought to be $\approx 4 M_{\odot}$.
emerging from a point at radius $r$ in the Schwarzschild geometry, given by
$$
\begin{equation*}
\sin \psi_{\mathrm{c}}=\frac{\sqrt{27} M}{r}\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{25.16}
\end{equation*}
$$

At angles greater than $\psi_{c}$, the photons spiral into the origin and are captured. Notice how the angle decreases to zero for $r=2 M$ : we cannot see any light signal from this point as all photons are captured.

The Schwarzschild metric is invariant with respect to time reversal, which means that this argument also applies to photons arriving at the location $r$ where the astronaut finds herself. This affects what the astronaut observes: photons from $r \leq 2 M$ cannot reach her since photons do not emerge from a black hole. For photons from elsewhere, $\psi_{\mathrm{c}}$ represents a limit to what the astronaut outside the hole can see, with incoming photons from angles less than $\psi_{c}$ being the only ones that she is able to detect. The rest of her field of vision is black: this is the astronaut 'seeing' the black hole.

\section*{Example 25.4}

For example, when $r=3 M$ we have $\sin \psi_{\mathrm{c}}=1$ and $\psi_{c}=\pi / 2$. The black hole then occupies exactly half of the sky. ${ }^{13}$ As the astronaut gets closer to the black hole, the hole takes up more and more of the sky, until, very close to the horizon, light from the rest of the universe is only experienced through a small cone.

The photons that do reach the astronaut from other stars have their energies increased (or blueshifted) compared to the energies of these photons at infinity. This can be seen by evaluating ${ }^{14}$ the energy measured by a stationary observer at radius $r$, which is
$$
\begin{equation*}
E_{\mathrm{obs}}=p^{\hat{t}}=\left(\boldsymbol{e}_{t}\right)^{\hat{t}} p^{t} \tag{25.17}
\end{equation*}
$$
with $\left(\boldsymbol{e}_{t}\right)^{\hat{t}}=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}}$. Recall that $p_{t}$ is conserved along the geodesics, so ${ }^{15} p^{t}(r)=\hbar \omega_{\infty}(1-2 M / r)^{-1}$. This yields
$$
\begin{equation*}
E_{\mathrm{obs}}=\hbar \omega_{\infty}\left(1-\frac{2 M}{r}\right)^{-\frac{1}{2}} \tag{25.18}
\end{equation*}
$$
which is simply eqn 22.16 for the gravitational redshift rearranged. The final thing the astronaut sees is a tiny chink of blue light. Bon voyage.

\subsection*{25.5 Gravitational collapse}

How are black holes created? When a main-sequence (i.e. a fairly average-sized) star exhausts its nuclear fuel, it generally expands to form a red giant, and subsequently collapses under its own gravitational attraction to form a white dwarf star. However, when a star of mass $\gtrsim 1.4 M_{\odot}$ uses up its nuclear fuel, ${ }^{16}$ then things can proceed differently.

If the star remains massive enough (i.e. it does not shed its outer layers of mass) it does not achieve the equilibrium state of a stable white dwarf. Instead, it will continue to collapse. If massive enough it contracts through the Schwarzschild radius $r=2 M$ as shown in Figs. 25.8 and 25.9. It then continues to collapse further until it is compressed to infinite density. That is, its mass is compressed to a point at the origin, with the result being a singularity in spacetime at $r=0$.

Can an observer ever hope to observe this singularity? Not at all. We saw in the last section that regular light signals from an in-falling astronaut become less frequent (and also highly redshifted) as the astronaut nears $r_{\mathrm{S}}$. This is also true of photons from the surface of a collapsing star. The image of the shrinking star appears to slow and darken. It freezes, becoming completely black in appearance at its outer surface approaches $r_{\mathrm{S}}=2 M$ from above. The light cones of the photons emitted from the surface of the star are shown in Figs. 25.8 and 25.9. Light emitted as the surface of the star passes through $r_{\mathrm{S}}$ has an ingoing wavefront that falls into the singularity at $r_{\mathrm{S}}=0$ and an outgoing wavefront that remains at the radius $r_{\mathrm{S}}=2 M$ for ever. Light emitted as the surface collapses further is drawn in towards the black hole, no matter whether it was directed towards larger or smaller $r$. (As explained in the next chapter, this curious situation leads to the formation of a so-called closed trapped surface.) At this point, the star itself is something of an irrelevance for the outside world which can no longer interact with it in any way. It is as if the star has now entered a Universe of its own. What matters for the outside Universe is now only the event horizon and the exterior geometry. By the same token, the singularity is hidden from the outside world.
Recall from Chapter 19 that we can map the presence of infinities and singularities in a spacetime using a Penrose diagram to depict the conformal structure. The conformal structure of spacetime that follows from the gravitational collapse that results in a black hole is shown in the Penrose diagram in Fig. 25.10. The black-hole singularity at $r=0$ is shown by the double line, along with the event horizon H at $r_{\mathrm{S}}=2 M$. The black-hole singularity is a spacelike surface (just as we had for the $t=0$ point in the Robertson-Walker Penrose diagram in Chapter 19.) The reason for this is the swapped roles of time and space inside the event horizon. This means we describe $r=0$ as a spacelike singularity.

For travellers on timelike geodesics that end up at $i^{+}$all of spacetime is visible, except the region for which $r<2 M$. For those unlucky enough to have fallen through the event horizon, all light rays (which, remember, travel at $45^{\circ}$ in these diagrams) will meet the singularity. There is a future horizon at the singularity: the observer in the region $r<2 M$ cannot access photons from the whole of the space. The Penrose diagram in Fig. 25.11 also shows the surface of a star as it collapses, passing though $r_{\mathrm{S}}$ and eventually meeting the singularity.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-119.jpg?height=304&width=380&top_left_y=407&top_left_x=1207)

Fig. 25.9 Gravitational collapse with one spatial dimension suppressed.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-119.jpg?height=430&width=295&top_left_y=786&top_left_x=1247)

Fig. 25.10 Penrose diagram showing the structure of spacetime containing a black hole with event horizon $H$.
![](https://cdn.mathpix.com/cropped/2025_04_12_c5e9e66267f04e703cbdg-119.jpg?height=402&width=305&top_left_y=1313&top_left_x=1279)

Fig. 25.11 Penrose diagram showing the gravitational collapse of a star (hatched area) to form a black hole.

\section*{Chapter summary}
- Black holes are regions of spacetime where gravitational curvature doesn't allow light to escape. The spacetime of spherically symmetric black holes is described by the Schwarzschild metric.
- The Schwarzschild radius $r_{\mathrm{S}}=2 M$ represents the event horizon for the black hole. Inside the event horizon, the light cone structure means that all matter will inevitably meet the singularity in spacetime at $r=0$.
- An astronaut will not notice anything special when passing the point $r=r_{\mathrm{S}}$ since this does not represent a physical singularity in spacetime. They will experience huge radially stretching gravitational forces as they approach $r=0$.
- Although it takes an astronaut a finite proper time to meet the singularity, the coordinate time diverges.

\section*{Exercises}
(25.1) Consider the Schwarzschild metric in the form
$$
\begin{equation*}
\mathrm{d} s^{2}=-\left(1-v^{2}\right) \mathrm{d} t^{2}+\left(1-v^{2}\right)^{-1} \mathrm{~d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{25.19}
\end{equation*}
$$
where $v^{2}=2 M / r$ and $\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi$.
(a) Compute the metric that results from a coordinate transformation $\mathrm{d} t=\mathrm{d} T-R(r) \mathrm{d} r$, where $R(r)$ is a function of $r$ only.
(b) What function $R(r)$ is required to give $g_{r r}=1$ ?
(c) Using the function from (b), show that the metric becomes
$$
\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\left[\mathrm{d} r+\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} T\right]^{2}+r^{2} \mathrm{~d} \Omega^{2}
$$
(25.20)
(d) Comment on the form of the metric describing a hypersurface of constant $T$.
(e) What are the velocities $\mathrm{d} r / \mathrm{d} \tau$ and $\mathrm{d} r / \mathrm{d} T$ for a radial plunge in these coordinates.
Hint: Make use of the radial plunge results from Chapter 22.
(f) How do light cones behave in these coordinates?
(g) Show that the coordinate speed for a falling observer is always less than the coordinate speed of light.

The coordinates used in eqn 25.20 are known as Painlevé-Gullstand coordinates or global rain coordinates and demonstrate that there is no physical singularity at $r=2 M$ (see the next chapter). The line element in these coordinates was independently proposed in 1922 by Paul Painlevé (twice Prime Minister of the French Third Republic) and Allvar Gullstrand (winner of the Nobel Prize in Physiology in 1911). Both were concerned that the solution to the Einstein equation they had discovered showed that relativity allowed infinite numbers of solutions, and so was incomplete. Lemaitre showed in 1933 that these newly discovered solutions were simply the results of a coordinate transformation of the Schwarzschild line element, as the problem demonstrates. Incidentally, Gullstand had also blocked Einstein from receiving the Nobel Prize in Physics for his formulation special relativity in 1905, as he believed that theory to be incorrect.
(25.2) We shall derive, and justify the name of, the global rain coordinates from the previous problem using an array of in-falling clocks. See the books by Taylor, Wheeler, and Bertschinger, and by Moore (whose approach we follow here) for more details.

The synchronized clocks are dropped at a steady rate from rest at infinity (where their readings coincide with the Schwarzschild coordinate time $t$ ), and fall radially into the black hole. Observers travel with the falling clocks and when they coincide with an event, they read off the time $\breve{t}$ and record the value of $r, \theta$ and $\phi$. To compute the metric line element in these coordinates we note that $\breve{t}=\breve{t}(r, t)$ and so
$$
\begin{equation*}
\mathrm{d} \breve{t}=\left(\frac{\partial \breve{t}}{\partial t}\right) \mathrm{d} t+\left(\frac{\partial \breve{t}}{\partial r}\right) \mathrm{d} r \tag{25.21}
\end{equation*}
$$
(a) Argue that, from the setup, we must have
$$
\begin{equation*}
\left(\frac{\partial \breve{t}}{\partial t}\right)=1 \tag{25.22}
\end{equation*}
$$

Next, we want to work out $\left(\frac{\partial \check{t}}{\partial r}\right)$, which we interpret as the difference in $\breve{t}$ measured by two clocks, separated by a distance $\mathrm{d} r$ evaluated a coordinate time $t$.
(b) Consider a pair of events separated by $\mathrm{d} r$ that occur at the same $t$. Show that for the observers attached to the clocks at these events
$$
\begin{equation*}
\frac{\mathrm{d} r}{\mathrm{~d} \tau}=-\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{25.23}
\end{equation*}
$$
and
$$
\begin{equation*}
\frac{\mathrm{d} r}{\mathrm{~d} t}=-\left(1-\frac{2 M}{r}\right)\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{25.24}
\end{equation*}
$$
(c) Call the time interval in $t$ between the clocks being dropped from infinity $\mathrm{d} t_{\mathrm{d}}$. Show that
$$
\begin{equation*}
\mathrm{d} t_{\mathrm{d}}=\frac{\mathrm{d} r}{(1-2 M / r)(2 M / r)^{\frac{1}{2}}} \tag{25.25}
\end{equation*}
$$
(d) Show that the proper time measured for a clock that falls radially from $r_{\mathrm{A}}$ to $r_{\mathrm{B}}$ is given by
$$
\begin{equation*}
\Delta \tau=\frac{2}{3} \frac{\left(r_{\mathrm{A}}^{\frac{3}{2}}-r_{\mathrm{B}}^{\frac{3}{2}}\right)}{(2 M)^{\frac{1}{2}}} \tag{25.26}
\end{equation*}
$$
(e) Explain why $\mathrm{d} \breve{t}=\mathrm{d} t_{\mathrm{d}}+\mathrm{d} \tau$.
(f) Show that
$$
\begin{equation*}
\frac{\partial \breve{t}}{\partial r}=\left(\frac{2 M}{r}\right)^{\frac{1}{2}}\left(1-\frac{2 M}{r}\right)^{-1} \tag{25.27}
\end{equation*}
$$
(25.3) Using the result of the previous problem, show that the metric for the global rain coordinates is
$$
\begin{aligned}
\mathrm{d} s^{2}= & -\left(1-\frac{2 M}{r}\right) \mathrm{d} \breve{t}^{2}+2\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} \breve{t} \mathrm{~d} r \\
& +\mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)
\end{aligned}
$$

How can this metric be used to argue that objects following timelike geodesics inside the event horizon must move inwards?
(25.4) A particle in the Schwarzschild geometry around a black hole starts in the $\theta=\pi / 2$ plane at radius $r$, moving purely tangentially with a local velocity of magnitude $v_{0}$, as measured by a stationary observer.
(a) Using the fact that the relative velocity $v$ for observers with velocities $\boldsymbol{u}$ and $\boldsymbol{v}$ is determined by $\boldsymbol{u} \cdot \boldsymbol{v}=-\gamma(v)$, determine the components of the particle's velocity $\boldsymbol{v}$ in Schwarzschild coordinates at $r$.
(b) Assuming the particle moves freely, what are its values of the constants of the motion $\tilde{E}$ and $\tilde{L}$ in terms of $v_{0}$ ?
(c) If $r=4 M$, will a value of $v_{0}=1 / \sqrt{2}$ be sufficient to prevent the particle falling into the hole? See Blennow and Ohlsson for a more complete discussion of this problem.

  1. \curvearrowright The material in this chapter provides some background to the Newtonian theory of gravitation. Readers eager to gravitation. Readers eager to eral relativity right away can skip to the next chapter.